EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC FUNCTIONS

Calculus for the Life Sciences nd Edition Greenwell SOLUTIONS MANUAL Full download at: https://testbankreal.com/download/calculus-for-the-life-sciences-nd-editiongreenwell-solutions-manual-/ Calculus for the Life Sciences nd Edition Greenwell TEST BANK Full download at: https://testbankreal.com/download/calculus-for-the-life-sciences-nd-editiongreenwell-test-bank-/ Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC FUNCTIONS.1 Exponential Functions 1. number of folds 1 3 4 5... 10... 50 y = 3 x reflected in the x-axis, reflected across the y-axis, and translated up units. This is graph A. 10. The graph of y = + 3 x is the same as the layers of paper 4 8 16 3... 104... 50 graph of x y = 3 x. This is the graph of 50 = 1.15899907 10 15 y = 3 reflected in the y-axis and translated. 500 sheets are inches high two units downward. This is graph B. 500 50 11. The graph of y = 3 x 1 is the graph of y = 3 x = in. x in. 50 x = 500 = 71, 079, 539.57 mi = 4.50359967 10 1 in. translated 1 unit to the right. This is graph C. 1. Some of the functions are equivalent to each other. y = 3 x 1 x 1 1 x = ( 3 ) (graph C), 3. The graph of y = 3 x is the graph of an x ( 1 ) y = 3 = 3 (graph D), and exponential function T h is is graph E. Copyright 015 Pearson Education, Inc. 89

y = a x with a > 1. y = 3 x +1 = 3(3) x (graph F). 13. x = 3 14. 4 x = 64 4. The graph of y = 3 x is the graph of y = 3 x x = 5 4 x = 4 3 reflected across the y-axis. This is graph D. x = 5 x = 3 5. The graph of y = ( 1 ) 1 x is the graph of 3 15. 3 x = 1 81 16. e x = 1 e 5 y = (3 1 ) 1 x or y = 3 x 1. This is the graph of y = 3 x translated 1 unit to the right. This is graph C. 17. 3 x = x 1 3 4 4 3 = 3 x = 4 e x = e 5 x = 5 4 x = 8 x +1 18. 5 x = 15 x + 6. The graph of y = 3 x +1 is the graph of y = 3 x ( ) x = ( 3 ) x +1 (5 ) x = (5 3 ) x + translated 1 unit to the left. This is graph F. x = 3x +3 5 x = 5 3x + 6 7. The graph of y = 3(3) x is the same as the x = 3x + 3 x = 3x + 6 x = 3 6 = x graph of y = 3 x +1. This is the graph of y = 3 x x = 3 translated 1 unit to the left. This is graph F. 3 x 8. The graph of y = ( 1 ) is the graph of ( 4 ) x +3 = ( 6 ) x 5 19. 16 x +3 = 64 x 5 0. (e 3 ) x = e x + 5 4 x +1 1 x 30 y = (3 1 ) x = 3 x. This is the graph of y = 3 x = 4 x + 1 = 1 x 30 reflected in the y-axis. This is graph D. 4 = 8x e 6 x = e x + 5 6 x = x + 5 9. The graph of y = 3 x is the same as the 1 = x 4 graph of y = 3 x +. This is the graph of Copyright 015 Pearson Education, Inc. 89

90 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 90 1. e x = (e 4 ) x + 3 e x = e 4 x +1 x = 4 x + 1 5x = 1 1 x = 5. x = 8 x = 3 7. x x + x 7 = 9 (3 3 ) x = (3 ) x + x 3 3x = 3 x + x 3x = x + x 0 = x x 0 = x(x 1) x = 0 or x 1 = 0 1 x = 3 x = 3 or x = 3 3. 5 x = 1 5 5 x = 5 x = x = or x 4 x x = 1 x 4 8. x = 0 or x = e x +5 x + 6 = 1 e x +5 x + 6 = e 0 x + 5x + 6 = 0 ( x + 3)( x + ) = 0 x + 3 = 0 or x + = 0 x = 3 or x = 4. = 16 4 x x = ( 4 ) x 4 4 x 4 x +16 x = x 4 x = 4 x + 16 x 16 = 0 ( x + 4)( x 4) = 0 x = 4 or x = 4 9. Graph of y = 5e x + 30. Graph of y = e x 3 5. 5 x + x = 1 5 x + x = 5 0 x + x = 0 x( x + 1) = 0 x = 0 or x = 0 or x + 1 = 0 x = 1 6. 8 x = x + 4 31. Graph of y = 3e + ( 3 ) x = x + 4 3x = x + 4 3x = x + 4 x Copyright 015 Pearson Education, Inc. 89

91 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 91 3x x 4 = 0 (3x 4)( x + 1) = 0 x = 4 or x = 1 3 3. Graph of y = 4e x / 1 33. Answers will vary. Copyright 015 Pearson Education, Inc. 89

9 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 9 34. 4 and 6 cannot be easily written as powers of the same base, so the equation 4 x = 6 cannot be solved using this approach. 35. When x > 0, 3 x > e x > x because the functions are increasing. When x < 0, x 1 1 3 x < e x < x < < 1, and 3 3 the functions are decreasing. The graphs below illustrate this. x x 38. f ( x) = 500 3t (a) After 1 hour: f (1) = 500 3(1) = 500 8 = 4000 bacteria (b) initially: f (0) = 500 3(0) = 500 1 = 500 bacteria (c) The bacteria double every 3t = 1 hour, or every 1 hour, or 0 minutes. 3 (d) When does f (t ) = 3, 000? 36. 37. f ( x) approaches e A(t) = 3100e 0.0166t (a) 1970: t = 10.7188. A(10) = 3100e (0.0166)(10) = 3100e 0.166 3659.78 The function gives a population of about 3660 million in 1970. This is very close to the actual population of about 3686 million. (b) 000: t = 40 A(40) = 3100e 0.0166(40) = 3100e 0.664 601.90 The function gives a population of about 60 million in 000. (c) 015 : t = 55 3, 000 = 500 3t 64 = 3t 6 = 3t 6 = 3t t = The number of bacteria will increase to 3,000 in hours. 39. (a) Hispanic population: h(t ) = 37.79(1.01) h(5) = 37.79(1.01) 5 41.93 The projected Hispanic population in 005 is 41.93 million, which is slightly less than the actual value of 4.69 million. (b) Asian population: h(t ) = 11.14(1.03) t t h(5) = 11.14(1.03) t 1.48 The projected Asian population in 005 is 1.48 million, which is very close to the A (55) = 3100e 0.0166(55) = 3100e 0.913 = 774.5 4 From the function, we estimate that the world population in 015 will be 775 million.

93 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 93 actual value of 1.69 million. (c) Annual Hispanic percent increase: 1.01 1 = 0.01 =.1% Annual Asian percent increase: 1.03 1 = 0.03 =.3% The Asian population is growing at a slightly faster rate. (d) Black population: b(t ) = 0.5116t + 35.43 b(5) = 0.5116(5) + 35.43 37.99 The projected Black population in 005 is 37.99 million, which is extremely close to the actual value of 37.91 million.

94 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 94 (e) Hispanic population: Double the actual 005 value is (4.69) = 85.38 million. (b) f (t ) = Ce kt f (6) = Ce k (6) 680.5 = Ce 6k C = 680.5 e 6k f (15) = Ce 15k 758.6 = Ce 15k C = 758.6 e 15k The doubling point is reached when t 39, or in the year 039. Asian population: The function passes through the points (6, 680.5) and (15, 758.6). Now solve for k. 680.5 758.6 = e 6k e 15k e 15k 758.6 = e 6k 680.5 Double the actual 005 value is (1.69) = 5.38 million. The doubling point is reached when t 36, or in the year 036. Black population: Double the actual 005 value is e 9k = 758.6 680.5 9k = ln 758.6 680.5 k = 1 ln 758.6 0.0107186 9 680.5 680.5 Substitute this value into C = solve for C. 680.5 C = 63.95 e 6(0.0107186) f (t) = 63.95e 0.0107t e 6k and 40. (a) (37.91) = 75.8 million. The doubling point is reached when t 79, or in the year 079. 41. (a) (c) t = 0 corresponds to 00 and t = 5 corresponds to 05. f (t) = 63.95e 0.0107t f (0) = 63.95e 0.0107( 0) 805.8 f (5) = 63.95e 0.0107( 5) 855.9 The demand for physicians in 00 will be about 805,800 and in 00 will be about 855,900. These are very close to the values in the data. (d) The exponential regression function is f (t ) = 631.81e 0.014t. This is close to the function found in part (b). The data appear to fit an exponential curve. The data appear to fit an exponential curve.

95 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 95 (b) The year 1963 corresponds to t = 0 and the year 006 corresponds to t = 43. f (t ) = f 0 a t f (0) = f 0 a 0 f (43) = f 0 a 43 487 = f 0 9789 = f 0 a 43 9789 f 0 = a 43 The function passes through the points (0, 487) and (43, 9789). Now solve for a. 487 = 9789 a 43 = 9789 a = 9789 a 43 487 487 1 43 a 1.073 Thus, f (t ) = 487 (1.073) t. (c) From part (b), we have a 1.073, so the number of breeding pairs is about 1.073 times the number of breeding pairs in the previous year. Therefore, the average annual percentage increase is about 7.%. (d) Using a TI-84 Plus, the exponential regression is f (t ) = 398.8 (1.076) t. 4. (a) t = 10 N = 100 exp {9.8901 exp exp (.540 0.167 (10)) } 1006 t = 0 N = 100 exp {9.8901 exp exp (.540 0.167 ( 0)) } 43, 500 t = 30 N = 100 exp {9.8901 exp exp (.540 0.167 (30)) } 1, 637, 000 t = 40 N = 100 exp {9.8901 exp exp (.540 0.167 ( 40)) } 1, 931, 000 t = 50 N = 100 exp {9.8901 exp exp (.540 0.167 (50)) } 1, 969, 000 (b) (c) The number of bacteria levels off at about,000,000. (c) (d) f (11) = 0.8454e 0.081(11) 8.341 The risk increases about 6.8% f (6) = 0.8454e 0.081(6).947 The risk for a man with a score of 6 is about.9%. f (14) = 0.8454e 0.081(14) 15.57 The risk increases by about 1.6%. 43. (a) The function fits the data closely. (e) (f) f (3) = 0.110e 0.49(3) 0.376 The risk for a woman with a score of 3 is about 0.4%. f (1) = 0.110e 0.49(1) 1.7983

96 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 96 (b) f (3) = 0.8454e 0.081(3) 1.578 The risk for a man with a score of 3 is about 1.6%. The risk increases by about 1.6%.

97 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 97 44. 45. C = (g) f (3) = 0.110e 0.49(6) 0.4665 The risk for a woman with a score of 6 is about 0.47%. f (15) = 0.110e 0.49(15) 3.5308 The risk increases by about 3.1% D a 46. (a) (b) The emissions appear to grow exponentially. f ( x) = f 0 a x f 0 = 534 Use the point (108, 8749) to find a. 8749 = 534a 108 a 108 = 6750 534 e bt e at 8749 V (a b ) ( ) a = 108 534 1.06 (a) At time t = 0, D a C = V (a b ) D a = V (a b ) D a = V (a b ) ( e b(0) e a(0) ) ( e 0 e 0 ) (0) = 0 This makes sense because no cortisone has been administered yet. (b) As a large amount of time passes, the concentration is going to approach zero. The longer the drug is in the body, the lower the concentration. (c) D = 500, a = 8.5, b = 0.09, V = 3700 500 8.5 0.09t C = e e 8.5t 3700 (8.5 0.09) ) 450 = 31,117 ( e 0.09t e 8.5t ) ( f ( x) = 534 (1.06) x (c) From part (b), we have a 1.06, so the amount of carbon dioxide emissions is about 1.06 times the amount in the previous year. Therefore, the average annual percentage increase is about.6%. (d) The doubling point is reached when x 135. Thus, the first year in which emissions equal or exceed that threshold is 035. 47. Q(t) = 1000(5 0.3t ) [0, ] by [0, 0.15] T h e m a x i m u m concentration occurs at about t = 0.54 hour.

98 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 98 (a) (b) Q s. (6 ) 8 = 1000(5 0.3t ) 10 1 00 = 5 0.3t 15 5 0. 5 3 = 5 0.3t 3(6 3 = 0.3t ) 10 = t 55 It will take 10 months to reduce the.1 substance to 8 grams. 89 = I n 6 m o n t h s, t h e r e w i l l b e a b o u t 5 5 g r a m

99 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 99 48. (a) When x = 0, P = 1013. When x = 10, 000, P = 65. First we fit P = ae kx. 1013 = ae 0 a = 1013 P = 1013e kx 65 = 1013e k (10,000) 65 = e 10,000k 1013 (b) P = 1013e ( 1.34 10 4 ) x is the best fit. 10, 000k = ln 65 (c) P(1500) = 1013e 1.34 10 4 (1500) 89 1013 65 ( ) ln 1013 4 k = 1.34 10 10, 000 ( 1.34 10 4 ) x Therefore P = 1013e. Next we fit P = mx + b. P(11, 000) = 1013e 1.34 10 4 (11,000) 3 We predict that the pressure at 1500 meters will be 89 millibars, and at 11,000 meters will be 3 millibars. (d) Using exponential regression, we obtain We use the points (0, 1013) and P = 1038(0.9998661) x which differs (10, 000, 65). m = 65 1013 = 0.0748 10, 000 0 b = 1013 Therefore P = 0.0748x + 1013. 1 Finally, we fit P =. ax + b 1 1013 = a(0) + b 1 b = 9.87 10 4 1013 1 P = 1 ax + 1013 1 65 = 10, 000a + 1 1013 1 1 = 10, 000a + 65 1013 10, 000a = 1 1 65 1013 1 1 a = 65 1013.79 10 7 10, 000 Therefore, 1 P = (.79 10 7 ) x + (9.87 10 4 ). slightly from the function found in part (b) which can be rewritten as 49. (a) y = mt + b P = 1013(0.9998660) x. b = 0.75 Use the points (0, 0.75) and (4, 1900) to find m. m = 1900 0.75 = 1899.75 79.155 4 0 4 y = 79.155t + 0.75 y = at + b b = 0.75 Use the point (4, 1900) to find a. 1900 = a(4) + 0.75 1899.95 = 576a a = 1899.75 3.98 576 y = 3.98t + 0.75 y = ab t a = 0.75 Use the point (4, 1900) to find b. 1900 = 0.75b 4 b 4 = 1900 0.75

100 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 100 b = 4 1900 1.445 75 y = 0.75(1.445) t

101 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 101 (b) Solve for b. 11 37, 016 ab = = b 10 4, 3 ab 37, 016 b = 10 4, 3 1.557 The function y = 0.75(1.445) t is the best fit. (c) x = 30 corresponds to 015. y = 0.75(1.445) 30 = 17193.65098 17, 00 (d) The regression function is y = 0.1787(1.441) t. This is close to the function in part (b). 50. (a) C = mt + b Use the points (1, 4,3) and (11, 37,016) to find m. m = 37, 016 4, 3 1, 694 = 11 1 10 = 1, 69.4 Use the point (1, 4,3) to find b. y = 1, 69.4t + b 4, 3 = 1, 69.4(1) + b (b) Now solve for a. ab = 4, 3 37, 016 a 10 4, 3 = 4, 3 4, 3 a = 19, 369.64 37, 016 10 4, 3 Thus, the exponential function is C = 19, 369.64 (1.557 ) t The exponential function C = 19, 369.64 (1.557 ) is the best fit. t b = 305.6 Thus, the linear function is (c) Using a TI-84 Plus, the exponential C = 1, 69.4t + 305.6 regression is f (t ) = 19, 59.86 (1.585) t. C = at + b Use the points (1, 4,3) and (11, 37,016) to find a and b. This is very close to the function in part (b). (d) The year 01 corresponds to x = 1. 4, 3 = a(1) + b 4, 3 = a + b 37, 016 = a(11) + b 37, 016 = 11a + b Using the linear function, we have C = 1, 69.4t + 305.6 Now solve the system 11a + b = 37, 016 a + b = 4, 3 10a = 1, 694 a = 1, 694 = 177.45 10 Now solve for b. a + b = 4, 3 b = 4, 3 177.45 =, 549.55 Thus, the quadratic function is C = 177.45t +, 549.55

10 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 10 C = 1, 69.4 (1) + 305.6 = 58, 85.4 Using the quadratic function, we have C = 177.45t +, 549.55 C = 177.45 (1) +, 549.55 = 77, 78.35 C = ab t Use the points (1, 4,3) and (11, 37,016)to find a and b. Using the exponential function, we have C = 19, 369.64 (1.557 ) t C = 19, 369.64 (1.557 ) 1 97, 68 Using the regression function, we have ( ) ( ) t f t = 19, 59.86 1.585 f (1) = 19, 59.86 (1.585) 1 304, 013 4, 3 = ab 1 = ab 37, 016 = ab 11 Using the linear function, the total world wind energy capacity will be about 58,85 MW in 01. (continued on next page)

103 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 103 (continued) Using the quadratic function, the total world wind energy capacity will be about m 5. A = P (1 + r ) tm, P = 6, 000, r = 0.06, t = 4 77,78 MW in 01. (a) annually, m = 1 Using the exponential function, the total world wind energy capacity will be about 97,68 MW in 01. Using the regression function, the total world wind energy capacity will be about 304,013 MW in 01. The quadratic function value of 77,78 is the value closest to the predicted value of 73,000. 51. A = P (1 + r ) tm, P = 10, 000, r = 0.04, t = 5 (a) annually, m = 1 Interest = $3.936.0 $6, 000 A = 10, 000 1 + 0.04 5(1) = $6936.0 1 (c) quarterly, m = 4 = 10, 000(1.04) 5 0.06 4(4) = $1,166.53 Interest = $1,166.53 $10, 000 = $166.53 (b) semiannually, m = A = 10, 000 1 + 0.04 5() = $6993.6 (d) monthly, m = 1 = 10, 000(1.0) 10 m 0.06 4(1) A = 6, 000 1 + 1 = 6, 000(1.06) 4 = $3, 84.40 Interest = $3, 84.40 $6, 000 = $684.40 (b) semiannually, m = A = 6, 000 1 + 0.06 = 6, 000(1.03) 8 = $3.936.0 4( ) A = 6, 000 1 + 4 = 6, 000(1.015) 16 = $3, 993.6 Interest = $3, 993.6 $6, 000 0.06 4(1 ) = $1,189.94 A = 6, 000 1 + 1 Interest = $1,189.94 $10, 000 = 6, 000(1.005) = $189.94 48 = $33, 03.7 (c) quarterly, m = 4 Interest = $33, 03.7 $6, 000 5(4) A = 10, 000 1 + 0.04 = $703.7 4 (e) continuously, t = 4 = 10, 000(1.01) 0 A = 6, 000e (0.06)(4) = $1, 01.90 = $33, 05.48 Interest = $1, 01.90 $10, 000 Interest = $33, 05.48 $6, 000 = $01.90 = $705.48 (d) monthly, m = 1 53. For 6% compounded annually for years,

104 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 104 A = 10, 000 1 + 0.04 5(1) A = 18, 000(1 + 0.06) = 18, 000(1.06) 1 = 0, 4.80 = 10, 000(1.003) 60 For 5.9% compounded monthly for years, = $1, 09.97 Interest = $1, 09.97 $10, 000 A = 18, 000 1 + 0.059 1 = $09.97 (e) continuously, t = 5 A = 10, 000e (0.04)(5) = $1, 14.03 Interest = $1, 14.03 $10, 000 = $14.03 1( ) 1.059 4 = 18, 000 1 = 0, 48.54 The 5.9% investment is better. The additional interest is $0, 48.54 $0, 4.80 = $3.74.

105 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 105 r tm 54. A = P 1 + m, P = 5000, A = $7500, t = 5 57. 100 = 500 1 + 4 (a) m = 1 7500 = 5000 r 5(1) 500 1 + 1/ 5 3 = (1 + r ) 5 3 1 = r 0.084 r 1 100 = 1 + r.4 = 1 + r 1 + = (.4) 4 r (14)(4) 4 r 56 4 1/ 56 56 1/ 56 4 + r = 4(.4) r = 4(.4) 1/ 56 4 (b) The interest rate is about 8.4% m = 4 r 0.0630 The required interest rate is 6.30%. 7500 = 5000 1 + r 5(4) 4 58. (a) 30, 000 = 10, 500 r (4)(1) 1 + 4 1/ 0 3 r 0 = 1 + 300 4 105 = r 48 1 + 4 r 3 1 = 1 + r = 300 4 4 105 1/ 48 1/ 0 3 1/ 48 4 1 r 300 4 + r = 4 105 55. A = Pe rt 0.08 = r The interest rate is about 8.%. 1/ 48 300 r = 4 4 105 r 0.0884 (a) r = 3% A = 10e 0.03(3) $10.94 (b) r = 4% A = 10e 0.04(3) $11.7 (b) The required interest rate is 8.84%. r 30, 000 = 10, 500 1 + 365 300 r 4380 = 1 + 1(365)

106 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.1 EXPONENTIAL FUNCTIONS FUNCTIONS 106 105 365 56. (c) r = 5% r 300 1 4380 A = 10e 0.05(3) $11.6 + = 105 1 P = $5, 000, r = 5.5% 365 300 1 4380 Use the formula for continuous 365 + r = 365 105 (a) (b) (c) compounding, A = Pe rt. t = 1 A = 5, 000e 0.055(1) = $6, 413.5 t = 5 A = 5, 000e 0.055(5) = $3, 913.7 t = 10 A = 5, 000e 0.055(10) = $43, 331.33 r = 365 300 1 4380 365 105 0.0875 The required interest rate is 8.75%.

99 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 99 59. y = (0.9) t A = P 1 + r tm = 1000 1 + k 8( 4) (a) t y 0 (0.9) 0 = 1 1 (0.9) 1 = 0.9 (0.9) 0.85 3 (0.9) 3 0.78 4 (0.9) 4 0.7 5 (0.9) 5 0.66 6 (0.9) 6 0.61 7 (0.9) 7 0.56 m 4 k 3 = 1000 1 + 4 This represents the amount he could have had from January 1, 000, to January 1, 008, at a rate of k per annum compounded quarterly, $03.76. So, 10 1 j k 1000 1 + 1 + = 1990.76 4 k 3 and 1000 1 + 4 = 03.76. (b) 8 (0.9) 8 0.51 1 + k 3 4 =.0376 9 (0.9) 9 0.47 10 (0.9) 10 0.43 1000 1 + 10 1 k 1/ 3 1 + = (.0376) 4 1 + k = 1.05 4 k = 0.05 4 k = 0.1 or 10% Substituting, we have (c) Let x = the cost of the house in 10 years. 1000 1 + j 1 + 0.1 = 1990.76 Then, 0.43x = 165, 000 x 383, 71. 4 In 10 years, the house will cost about j 10 $384,000. 1 (1.05) = 1990.76 (d) Let x = the cost of the book in 8 years. j 10 Then, 0.51x = 50 x 98. + = In 8 years, the textbook will cost about 1 1.480 j 1/10 60. $98. A = P r 1 + tm m A = 1000 1 + j 5( ) = 1000 1 + j 10 1 + = (1.480) 1 + j = 1.04 j = 0.04

100 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 100 j = 0.08 or 8% This represents the amount in Bank X on The ratio k = 0.1 = 1.5, is choice a. January 1, 005. j 0.08 A = P 1 + r tm m. Logarithmic Functions j 10 3(4) k 1. 5 3 = 15 = 1000 1 + 1 + y 4 Since a = x means y = log a x, the equation j 10 k 1 in logarithmic form is log 5 15 = 3. = 1000 1 + 1 + 4. 7 = 49 This represents the amount in Bank Y on January 1, 008, $1990.76. Since a y = x means y = log a x, the equation in logarithmic form is log 7 49 =.

101 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 101 3. 3 4 = 81 1. log 0.001 = 3 Since a y = x means y = log a x, the equation log 10 0.001 = 3 in logarithmic form is log 3 81 = 4. 10 3 = 0.001 When no base is written, log 10 x is 4. 7 = 18 Since a y = x means y = log a x, the equation understood. 13. Let log 8 64 = x. Then, in logarithmic form is log 18 = 7. 5. 3 = 1 9 8 x = 64 8 x = 8 x = Thus, log 8 64 =. 14. Let log 9 81 = x. Then, Since a y = x means y = log a x, the equation 1 9 x = 81 9 x = 9 x = Thus log 9 81 =. in logarithmic form is log 3 9 =. 5 6. 4 15. log 4 64 = x 4 x = 64 4 x = 4 3 x = 3 = 16 5 16. log 3 7 = x 3 x = 7 3 x = 3 3 x = 3 Since a y = x means y = log a x, the equation 1 1 x 16 17. log = x = x = 4 x = 4 in logarithmic form is log 5/4 =. 16 16 5 7. log 3 = 5 18. log 3 1 = x 3 x = 1 3 x = 3 4 x = 4 81 81 Since y = log a x means a y = x, the equation 1 x 1 1/3 log = x = in exponential form is 5 = 3. 3 19. 4 4 1/3 8. log 3 81 = 4 x = 1 x = /3 x = 3 Since y = log a x means a y = x, the equation in exponential form is 3 4 = 81. 9. ln 1 = 1 e Since y = ln x = log e x means e y = x, the equation in exponential form is e 1 = 1. 0. log 4 1 = x 8 1 1 1/ 4 8 x = 4 = ( 3 ) x = 1/ 4 1 1

10 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 10 10. e 3x = x = 4 1 log 8 1 = 3 Since y = log a x means a y = x, the equation 1. ln e = x Recall that ln y means log e y. e x = e x = 1 11. in exponential form is 3 = 1. 8 log 100, 000 = 5 log 10 100, 000 = 5 10 5 = 100, 000 When no base is written, log 10 x is understood.. 3. ln e 3 = x Recall that ln y means log e y. e x = e 3 x = 3 ln e 5/3 = x e x = e 5/3 x = 5 3 4. ln ln 1 = x e x = 1 e x = e 0 x = 0 5. The logarithm to the base 3 of 4 is written log 3 4. The subscript denotes the base.

103 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 103 6. Answers will vary. 7. log 5 (3k ) = log 5 3 + log 5 k 38. log 1 10 = ln 10 ln 1.15 8. 9. log log 9 (4m) = log 9 4 + log 9 m 39. log 0.95 = ln 0.95 0.81 3 p = log 3 p log 5k 3 5k 3 3 = (log 3 3 + log 3 p) (log 3 5 + log 3 k ) = 1 + log 3 p log 3 5 log 3 k 15 p 41. log x 36 = x = 36 40. 1. ln 1. log 0.1 = ln 0.1.059.8 ln.8 4. log 9 7 = m 9 m = 7 30. log 7 7 y 31. = log 7 15 p log 7 7 y = (log 7 15 + log 7 p) (log 7 7 + log 7 y) = log 7 15 + log 7 p log 7 7 log 7 y = log 7 15 + log 7 p 1 log 7 y ln 3 5 = ln 3 5 ln 3 6 43. ( x ) 1/ = 36 1/ (3 ) m = 3 3 x = 1 6 log 16 = z 44. 3 m = 3 3 m = 3 m = 3 log 8 = 3 3 6 = ln 3 5 1/ ln 6 1/3 8 8 z = 16 y 4 y 3/ 4 = 8 3. ln = ln 3 + ln 5 1/ ln 6 1/ 3 ( 3 ) z = 4 ( y 3/ 4 ) 4/3 = 8 4/3 = ln 3 + 1 ln 5 1 ln 6 9 3 5 = ln 9 3 5 ln 4 3 4 3 3 3 z = 4 y = (8 1/3 ) 4 3z = 4 z = 4 3 y = 4 y = 16 = ln 9 5 1/3 ln 3 1/ 4 45. log 5 = 1 r 46. log 4 (5x + 1) = = ln 9 + ln 5 1/3 ln 3 1/ 4 r 1/ = 5 4 = 5x + 1 33. = ln 9 + 1 ln 5 1 ln 3 3 4 log b 3 = log b 5 = 5 log b = 5a 47. (r 1/ ) = 5 r = 5 log 5 (9 x 4) = 1 16 = 5x + 1 5x = 15 x = 3 34. log b 18 = log b ( 9) = log b ( 3 ) 5 1 = 9 x 4 35. = log b + log b 3 = log b + log b 3 = a + c log b 7b = log b 7 + log b b = log b 7 + 1 48. 9 = 9 x 1 = x log 4 x log 4 ( x + 3) = 1 x log = 1 4 x + 3

104 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 104 = log b ( 3 3 3 ) + 1 = log b 3 + log b 3 + 1 = 3 log b + log b 3 + 1 = 3a + c + 1 36. log b (9b ) = log b 9 + log b b = log b 3 + log b b = log b 3 + log b b = c + (1) = c + x = x + 3 x 1 = x + 3 4 4 1 4 x = x + 3 3x = 3 x = 1 ln 30 3.401 37. log 5 30 = ln 5.113 1.6094

105 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 105 49. log 9 m log 9 (m 4) = m 53. log ( x 1) log ( x + 1) = 50. 51. log 9 m 4 = 9 = m m 4 1 = m 81 m 4 m 4 = 81m 4 = 80m 0.05 = m This value is not possible since log 9 ( 0.05) does not exist. Thus, there is no solution to the original equation. log ( x + 5) + log ( x + ) = 1 log [( x + 5)( x + )] = 1 ( x + 5)( x + ) = 10 1 x + 7 x + 10 = 10 x + 7 x = 0 x( x + 7) = 0 x = 0 or x = 7 x = 7 is not a solution of the original equation because if x = 7, x + 5 and x + would be negative, and log ( ) and log ( 5) do not exist. Therefore, x = 0. log 3 ( x ) + log 3 ( x + 6) = log 3 [( x )( x + 6)] = ( x )( x + 6) = 3 x + 4 x 1 = 9 x + 4 x 1 = 0 54. log x 1 = x + 1 x 1 = 1 x + 1 x 1 4 = x + 1 4 x + 4 = x 1 x 4 x 5 = 0 ( x 5) ( x + 1) = 0 x = 5 or x = 1 x = 1 is not a solution of the original equation because if x = 1, x + 1 would not exist. Therefore, x = 5. ln(5x + 4) = 5x + 4 = e 55. ln x + ln 3x = 1 ln 3x = 1 3x = e 1 5x = e 4 e 4 x = 0.6778 5 x = e 3 e 1 1 x = = 0.350 3 3e 5. ( x + 7)( x 3) = 0 x = 7 or x = 3 56. ln( x + 1) ln x = 1 x + 1 x = 7 is not a solution of the original ln = 1 x equation because if x = 7, x + 6 would be negative, and log ( 1)does not exist. Therefore, x = 3. log 3 ( x + 17) log 3 ( x + 5) = 1 x + 17 x + 1 = e 1 x x + 1 = ex ex x = 1 x(e 1) = 1 1 log 3 = 1 x + 5 3 1 = x + 17 x = 0.580 e 1 x + 5

106 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 106 3x + 15 = x + 17 0 = x 3x + 0 = ( x 1)( x ) 57. x = 6 ln x = ln 6 x ln = ln 6 x = ln 6.5850 x = 1 or x = ln 58. 5 x = 1 x log 5 = log 1 x = log 1 1.5440 log 5

107 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 107 59. e k 1 = 6 ln e k 1 = ln 6 For exercises 65 68, use the formula a x = e (ln a ) x. (k 1) ln e = ln 6 k 1 = ln 6 ln e k 1 = ln 6 1 k = 1 + ln 6.7918 65. 66. 67. 68. 10 x +1 (ln 10)( x +1) = e 10 x = e (ln10) x e 3x = (e 3 ) x 0.09 x e 4 x = (e 4 ) x 0.0183 x 60. e y = 15 ln e y = ln 15 y ln e = ln 15 y (1) = ln 15 69. f ( x) = log (5 x) 5 x > 0 x > 5 x < 5 The domain of f is x < 5 or (, 5). y = ln 15 1.3540 70. f ( x) = ln ( x 9) Since the domain of f ( x) = ln x is (0, ), 61. 3 x +1 = 5 x ln 3 x +1 = ln 5 x ( x + 1) ln 3 = x ln 5 x ln 3 + ln 3 = x ln 5 x ln 5 x ln 3 = ln 3 x(ln 5 ln 3) = ln 3 ln 3 x =.1507 ln(5/3) 6. x +1 = 6 x 1 ln x +1 = ln 6 x 1 ( x + 1) ln = ( x 1) ln 6 x ln + ln = x ln 6 ln 6 x ln 6 x ln = ln + ln 6 x(ln 6 ln ) = ln + ln 6 x ln 3 = ln 1 x = ln 1.619 ln 3 the domain of f ( x) = ln ( x 9) is the set of all real numbers x for which x 9 > 0. To solve this quadratic inequality, first solve the corresponding quadratic equation. x 9 = 0 ( x + 3)( x 3) = 0 x + 3 = 0 or x 3 = 0 x = 3 or x = 3 These two solutions determine three intervals on the number line: (, 3), ( 3, 3), and (3, ). If x = 4, ( 4 + 3) ( 4 3) > 0. If x = 0, (0 + 3)(0 3) > 0. If x = 4, (4 + 3) (4 3) > 0. The domain is x < 3 or x > 3, which is written in interval notation as (, 3) (3, ). 63. 5 (0.10) x = 4(0.1) x 71. log A log B = 0 A ln [5 (0.10) x ] = ln [4 (0.1) x ] ln 5 + x ln 0.10 = ln 4 + x ln 0.1 x (ln 0.1 ln 0.10) = ln 5 ln 4 ln 5 ln 4 ln 1.5 x = = ln 0.1 ln 0.10 ln 1. log = 0 B A 0 = 10 = 1 B A = B A B = 0

108 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 108 1.39 Thus, solving log A log B = 0 is equivalent 64. 1.5 (1.05) x = (1.01) x to solving A B = 0 ln [1.5 (1.05) x ] = ln [ (1.01) x ] ln 1.5 + x ln 1.05 = ln + x ln 1.01 x(ln 1.01 ln 1.05) = ln 1.5 ln ln 0.75 x = ln (1.01 / 1.05) 7.4069 7. (log( x + )) log( x + ) (log( x + )) = (log( x + ))(log( x + )) 1og( x + ) (log x + log ) 1og( x + ) = log( x + ) log 100 log 0.30103 because 10 0.30103 =

109 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 109 73. Let m = log a x, n = log a x, and p = log a y. y y Then a m = x, a n = x, and a p = y. Substituting gives a m = x = a = a n p. y a p So m = n p. Therefore, log x a = log a x log a y. y 74. Let m = log a x r and n = log a x. n (b) h(t) = 11.14(1.03) t Double the 005 population is (1.69) = 5.38 million. 5.38 = 11.14(1.03) t 5.38 = (1.03) t 11.14 log 5.38 1.03 = t 11.14 = ( ) ln 5.38 t 11.14 ln 36.1 1.03 Then, a m = x r and a n = x. Substituting The Asian population is estimated to 75. (a) gives a m = x r = (a n ) r = a nr. Therefore, m = nr, or log a x r = r log a x. (b) ln t = 3.4 years ln (1 + 0.03) t = ln ln (1 + 0.06) 11.9 years (c) ln t = 9.0 years ln (1 + 0.08) ln 76. (a) t = 1 ln (1 + 0.06) It will take about 1 years for the population to at least double. double their 005 population in 036. 78. Double the 006 population is (9789) = 19,578. 19, 578 = 487 (1.073) x 19, 578 = (1.073) x 487 ln 19, 578 = x ln 1.073 487 ln 19, 578 487 x = 5.9 ln 1.073 The number of bald eagle breeding pairs will double the 006 level in 1963 + 53 = 016. 79. Set the exponential growth functions y = 4500 (1.04) t and y = 3000 (1.06) t equal (b) 77. (a) ln 3 t = 19 ln (1 + 0.06) It will take about 19 years for the population to at least triple. h(t) = 37.79(1.01) t Double the 005 population is (4.69) = 85.38 million. 85.38 = 37.79(1.01) t 85.38 = (1.01) t 37.79 to each other and solve for t. 4500 (1.04) t = 3000 (1.06) t 1.5 (1.04) t = 1.06 t ln ( 1.5 (1.04)t ) = ln (1.06) t ln 1.5 + t ln1.04 = t ln1.06 ln 1.5 = t (ln1.06 ln1.04) ln 1.5 t = 1.9 ln 1.06 ln 1.04 After about 1.3 years, the black squirrels will outnumber the gray squirrels. Verify this with

110 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 110 85.38 log 1.01 37.79 = t ( ) a graphing calculator. ln 85.38 t 37.79 = ln 1.01 39. The Hispanic population is estimated to double their 005 population in 039. [0, 40] by [0, 0,000] 80. (a) New York: f (t ) = 18. (1.001) t Florida: f (t ) = 14.0 (1.017 ) t

111 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 111 (b) 18. (1.001) t = 14.0 (1.017 ) t (c) Yes, ln 0.693. 18. (1.017 ) t 1.017 t 8. H = [P 1 ln P 1 + P ln P = = + P ln P + P ln P ] 14.0 (1.001) t 1.001 3 3 4 4 H = [0.51 ln 0.51 + 0.34 ln 0.34 ln 18. 14.0 = t ln 1.017 1.001 ln 18. t 14.0 = 16.54 + 0.081 ln 0.081 + 0.074 ln 0.074] H = 1.101 83. (a) 3 species, 1 3 each: ln 1.017 P = P = P = 1 1 3 1.001 3 If this trend continued, then the population of Florida would have exceeded that of New York about 16.5 years after 1994, or sometime in 011. H = ( P 1 ln P 1 + P ln P + P 3 ln P 3 ) = 3 1 1 1 ln = ln 3 3 3 1.099 (b) 4 species, 1 4 each: 1 P 1 = P = P 3 = P 4 = 4 H = (P 1 ln P 1 + P ln P + P 3 ln P 3 + P 4 ln P 4 ) = 4 1 ln 1 = ln 1 (c) New York: f (t) = 19.5 (1.004) t 4 4 4 1.386 Florida: f (t ) = 19.1(1.01) t (c) Notice that 19.5 (1.004) t = 19.1(1.01 ) t ln 1 = ln (3 1 ) 1 = ln 3 1.099 and (1.004) t t 1.004 = (1.01) t 1.01 = 19.1 19.5 t ln 1.004 1.01 = ln 19.1 19.5 ln 19.1 t 19.5 = ln 1.004.61 84. 1.01 If this trend continued, then the 3 ln 1 = ln (4 1 ) 1 = ln 4 1.386 4 by Property c of logarithms, so the populations are at a maximum index of diversity. mx + N = m log b x + log b n p o = log b x m + log b n = log b nx m pulation of Florida would have exceeded that of New York about.6

11 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 11 years after 011, or sometime in 014. (d) It is less accurate to predict events far beyond the given data set. 81. (a) The total number of individuals in the community is 50 + 50, or 100. Let P = 50 1 = 0.5, P = 0.5. 100 H = 1[P 1 ln P 1 + P ln P ] = 1[0.5 ln 0.5 + 0.5 ln 0.5] 0.693 (b) For species, the maximum diversity is ln. 85. = log b y = Y Thus, Y = mx + N. 0.8168 0.0154 log w A = 4.688w (a) (b) 0.8168 0.0154 log 4000 A = 4.688 (4000) 590 cm A = 4.688 (8000) 0.8168 0.0154 log 8000 411 cm

113 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 113 (c) Graph the equations Y 1 = 4.688x 0.8168 0.0154 log x and Y = 4000, then find the intersection to solve 4000 = 4.688x 0.8168 0.0154 log w. (b) V ln = ln (1 + r ) V 1 t t 1 ln V ln V 1 = (t t 1 ) ln (1 + r ) ln V ln V 1 = ln (1 + r ) t t 1 [0, 10,000] by [0, 5000] An infant with a surface area of 4000 (c) Substitute ln (1 + r ) = ln V ln V 1 t t 1 ln t =, then simplify. ln (1 + r ) ln ln t = = into square cm weighs about 7430 g. 86. If the number N is proportional to m 0.6, where m is the mass, then N = km 0.6, for some constant of proportionality k. Taking the common log of both sides, we have log N = log(km 0.6 ) = log k + log m 0.6 = log k 0.6 log m. ln (1 + r ) ln V ln V 1 t t 1 = ( t t 1 ) ln ln V ln V 1 (d) Use the formula derived in part (c) with t 1 = 0, t = 4.5, and V = 1.55V 1. ( 4.5 0 ) ln 4.5 ln t = = ln (1.55V 1 ) ln V 1 1.55V ln 1 This is a linear equation in log m. Its graph is a straight line with slope 0.6 and vertical intercept log k. = 4.5 ln 7.1 ln 1.55 V 1 87. C (t ) = C 0 e kt The doubling time for the tumor is about 7.1 years. When t = 0, C(t) =, and when t = 3, C(t) = 1. 89. (a) y (t ) = y 0 e kt = C 0 e k (0) y (t ) = e kt C 0 = y 0 1 = e 3k y (t) ln = kt 1 = e 3k y 0 0 1 y (t) ln = k, which is a constant. 3k = ln 1 = ln 1 = ln t y k = ln 3 T = 1 ln C T = k C 1 1 5C ln 1 = 3 ln 5 7.0

114 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 114 (b) The data in Section.1, exercise 40 is as follows: ln 3 0 C 1 t ln The drug should be given about every 7 hours. 88. (a) V 1 = V 0 (1 + r ) 1 V = V (1 + r ) t t t t V V 0 (1 + r ) (1 + r ) = = = 1 + r 1 t Year ( ) t t t = 19) Demand for Physicians (in thousands) 006 680.5 015 758.6 00 805.8 05 859.3 Evaluate the expression for the years 015 (t = 9), 00 (t = 14), and 05 V 1 V 0 (1 + r ) 1 (1 + r ) 1 (continued on next page)

115 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 115 (continued) 015: 1 ln 758.6 0.0107 9 680.5 00: 1 ln 805.8 0.0107 14 680.5 05: 1 ln 859.3 0.018 19 680.5 90. (a) From the given graph, when x = 10 g, y 1.3 cm 3 / g / hr, and when x = 1000 g, y 0.41 cm 3 / g / hr. (b) If y = ax b, then (c) ln y = ln (ax b ) = ln a + b ln x. Thus, there is a linear relationship between ln y and ln x. 1.3 = a(10) b (b) Yes, the graph appears to be more linear, especially if the first point is eliminated. (c) Using a TI-84 Plus, the least squares line for the data in part (b) is Y = 0.0940 x + 9.348. (d) Using a TI-84 Plus, the exponential regression function for the data in part (a) x is Y = 11, 471(1.098). (e) ln Y = ln ( 11, 471(1.098) x ) = ln 11, 471 + ln (1.098 x ) 0.41 = a(1000) b = ln 11, 471 + x ln 1.098 1.3 a(10) b 1.3 10 b 9.348 + 0.940 x = = 0.41 a(1000) b 0.41 1000 9. For concert pitch A, f = 440. ln 1.3 = ln (0.01) b 0.41 ln 1.3 = b ln (0.01) 0.41 1.3 ( ) ln 0.41 b = ln (0.01) 0.5 93. P = 69 + 1 log (440/440) = 69 + 1 log (1) = 69 + 1 0 = 69 For one octave higher than concert pitch A, f = 880. P = 69 + 1 log (880/ 440) = 69 + 1 log () = 69 + 1 1 = 69 + 1 = 81 N (r ) = 5000 ln r Substituting this value into 1.3 = a(10) b, 1.3 = a(10) 0.5 1.3 a =.3. (10) 0.5 (a) (b) (c) N (0.9) = 5000 ln (0.9) 530 N (0.5) = 5000 ln (0.5) 3500 N (0.3) = 5000 ln (0.3) 6000 Therefore, y =.3x 0.5. (d) If x = 100, y =.3(100) 0.5 0.73. (d) N (0.7) = 5000 ln (0.7) 1800 About 1800 years have elapsed since the split if 70% of the words of the ancestral

116 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 116 91. (a) We predict that the basal metabolism for a marsupial carnivore with a body mass of 100 g will be about 0.73 cm 3 /g / hr. (e) language are common to both languages today. 5000 ln r = 1000 ln r = 1000 5000 1 ln r = 5 r = e 1/ 5 0.8 No, the data do not appear to lie along a straight line.

117 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 117 94. Decibel rating: 10 log I I 0 10 log I = 75 I 0 I (a) Intensity, I = 115I 0 log I 0 = 7.5 115I 0 10 log = 10 log115 1 I 0 (b) I = 9,500,000I 0 6 9.5 10 I 0 6 10 log = 10 log 9.5 10 I 0 70 (c) I = 1,00,000,000I 0 1. 10 9 I 10 log = 10 log 1. 10 0 9 log I log I 0 = 7.5 log I 0 = log I 7.5 Substitute for I 0 in the equation for log I 1. log I 1 = 8.5 + log I 0 = 8.5 + log I 7.5 = 1 + log I log I 1 log I = 1 I I 1 log = 1 I (d) Then 1 = 10, so I = 1 I 1. This means the I 0 I 10 91 intensity of the sound that had a rating of 75 I = 895,000,000,000I 0 1 decibels is 10 as intense as the sound that had a rating of 85 decibels. 8.95 10 1 1 I 0 11 10 log = 10 log 8.95 10 I 0 10 96. R(I ) = log I I 0 R(1, 000, 000 I ) = log 0 1, 000, 000 I (e) I = 109,000,000,000,000I 0 14 (a) 0 I 0 1.09 10 I 0 14 10 log = 10 log 1.09 10 = log 1, 000, 000 = 6 I 0 140 (b) R(1, 000, 000, 000 I 0 ) log 1, 000, 000, 000 I 0 (f) I 0 = 0.000 microbars = I 1, 00, 000, 000I 0 = 1, 00, 000, 000(0.000) = 40, 000 microbars 895, 000, 000, 000I 0 = 895, 000, 000, 000(0.000) = 179, 000, 000 microbars (c) = log 100, 000, 000 = 8 R(I ) = log I I 0 6.7 = log I I 0 0

118 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section. LOGARITHMIC FUNCTIONS FUNCTIONS 118 95. Let I 1 be the intensity of the sound whose decibel rating is 85. 10 log I 1 = 85 I 0 log I 1 = 8.5 I 0 log I 1 log I 0 = 8.5 log I 1 = 8.5 + log I 0 Let I be the intensity of the sound whose decibel rating is 75. (d) (e) I 10 6.7 = I 5, 000, 000I 0 I 0 R( I ) = log I I 0 I 8.1 = log I 0 I 10 8.1 = I 16, 000, 000I 0 I 0 1985 quake = 16, 000, 000 I 0 5 1999 quake 5, 000, 000I 0 The 1985 earthquake had an amplitude more than 5 times that of the 1999 earthquake.

109 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 109 (f ) R( E) = log E (b) For laundry solution: 3 E 0 11 = log[h + ] 10 11 = [H + ] For the 1999 earthquake 6.7 = log E 10.05 = log E For black coffee: 5 = log[h + ] E E 0 = 10 3 E 0 E 0 10.05 10.05 For the 1985 earthquake, 8.1 = log E 3 E 0 1.15 = log E E 0 E = 10 E 0 10 10 5 = [H + ] 5 11 = 10 6 = 1, 000, 000 10 The coffee has a hydrogen ion concentration 1,000,000 times greater than the laundry mixture..3 Applications: Growth and Decay E 1.15 1.15 = 10 E 0 E = 10 E 0 1. y0 represents the initial quantity; k represents The ratio of their energies is 10 1.15 E 0.1 10 10.05 = 10 16 E 0 The 1985 earthquake had an energy about 16 times that of the 1999 earthquake. (g) Find the energy of a magnitude 6.7 earthquake. From part f, we have the rate of growth or decay.. k is positive in the exponential growth function. k is negative in the exponential decay function. 3. The half-life of a quantity is the time period for the quantity to decay to one-half of the initial amount. E = E 0 10 10.05. For an earthquake that 4. Assume that y = y 0 e kt represents the amount releases 15 times this much energy, E = E 0 (15)10 10.05. R( E (15)10 10.05 ) = log E 0 (1 5)10 10.05 remaining of a radioactive substance decaying with a half-life of T. Since y = y 0 is the amount of the substance at time t = 0, y 0 then 0 y = is the amount at time t = T. 3 E 0 = log (15 10 10.05 ) 7.5 3 So, it s true that a magnitude 7.5 earthquake releases 15 times more energy than one of magnitude 6.7. 97. ph = log[h + ] Therefore, y 0 = y 0 e kt. Solving for k yields 1 = e kt ln 1 = kt

110 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 110 (a) For pure water: ln ( 1 ) ln ( 1 ) ln 7 = log[h + ] 7 = log[h + ] 10 7 = [H + ] k = = T T =. T For acid rain: 5. Assume that y = y 0 e kt is the amount left of a 4 = log[h + ] radioactive substance decaying with a half-life 4 = log[h + ] 10 4 = [H + ] of T. From Exercise 4, we know k so ln = T, 10 4 3 = 10 = 1000 0 10 7 y = y 0 e ( ln /T )t = y 0 e (t /T ) ln = y e t /T t /T ( ln ) 1 t /T The acid rain has a hydrogen ion = y 0 t /T 1 1 = y 0 = y 0 concentration 1000 times greater than pure water.

111 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 111 6. (a) P = P 0 e kt When t = 1650, P = 500. When t = 010, P = 6756. 500 = P 0 e 1650k 6756 = P 0 e 010k 6756 P 0 e = 010k For 15 days, t = 15 4 or 360. y = (1) 360/1 = 30 = 1, 073, 741, 84 8. y = y 0 e kt y = 40, 000, y 0 = 5, 000, t = 10 (a) 40, 000 = 5, 000e k (10) 500 P 0 e 1650k 1.6 = e 10k 6756 = e 360k 500 360k = ln 6756 500 k = ( ) ln 6756 500 360 0.0073 ln 1.6 = 10k 0.047 k The equation is y = 5, 000e 0.047t. (b) y = 5, 000e 0.047t = 5, 000(e 0.047 ) t Substitute this value (include all decimal places) into 500 = P 0 e 1650k to find P 0. 500 = P 0 e 1650(0.0073) (c) = 5, 000(1.048) t y = 60, 000 60, 000 = 5, 000e 0.047t.4 = e 0.047t 500 P 0 = 0.00385 ln.4 = 0.047t e 1650(0.0073) 18.6 t (b) Therefore, P(t) = 0.00385e 0.0073t. P(1) = 0.00386e 0.0073 0.0033 million, or 3300. 9. y = y 0 e kt There will be 60,000 bacteria in about 18.6 hours. 7. (a) The exponential equation gives a world population of only 3300 in the year 1. (c) No, the answer in part (b) is too small. Exponential growth does not accurately describe population growth for the world over a long period of time. y = y 0 after 1 hours. (a) y = 0, 000, y 0 = 50, 000, t = 9 0, 000 = 50, 000e 9k 0.40 = e 9k ln 0.4 = 9k 0.10 k The equation is y = 50, 000e 0.10t. y = y 0 e kt y 0 = y 0 e 1k = e 1k ln = ln e 1k 1k = ln (b) 1 (50, 000) = 5, 000 k = l n 0.05776 1 y = y 0 e 0.05776t

Section.3 APPLICATIONS: GROWTH AND DECAY 11 11 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC FUNCTIONS 10. 5, 0 0 0 = 5 0, 0 0 0 e 0. 1 0 t 0. 5 = e 0. 1 0 t l n 0. 5 = 0. 1 0 t 6. 8 t Half the bacteria remain after about 6.8 hours. f (t ) = 500 e 0.1t

113 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 113 (b) (ln /1)t y = y 0 e = y 0 e (ln )(t /1) ln t /1 (a) f (t ) = 3000 3000 = 500e 0.1t = y 0 [e ] 6 = e 0.1t = y 0 t /1 since e ln = ln 6 = 0.1t t 17.9 It will take 17.9 days. (c) For 10 days, t = 10 4 or 40. y = (1) 40/1 = 0 = 1, 048, 576

114 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 114 (b) If t = 0 corresponds to January 1, the date January 17 should be placed on the product. January 18 would be more than 17.9 days. n (e) The results are summarized in the following table. Value of k for [0, 35] Value of k for [4, 49] 11. Use y = y 0 e kt. 0.00093 0.0017 When t = 5, y = 0.37 y 0. 0.37 y 0 = y 0 e 5k 0.37 = e 5k 5k = ln(0.37) k = ln(0.37) 0.1989 5 1. (a) From the graph, the risks of chromosomal abnormality per 1000 at ages 0, 35, 4, and 49 are, 5, 4, and 15, respectively. (b) (Note: It is difficult to read the graph accurately. If you read different values from the graph, your answers to parts (b) (e) may differ from those given here.) y = Ce kt When t = 0, y =, and when t = 35, y = 5. = Ce 0k 5 = Ce 35k 13. 14. 3.3 10 5 5.5 10 4 6.3 10 7 A(43, 000) = A 0 4.1 10 7 The value of n should be somewhere between 3 and 4. 1 t /5600 A(t ) = A 0 1 43,000/5600 0.005 A 0 About 0.5% of the original carbon 14 was present. P(t) = 100e 0.1t (a) (b) P(4) = 100e 0.1(4) 67% P(10) = 100e 0.1(10) 37% 5 Ce 35k (c) 10 = 100e 0.1t = Ce 0k 0.1 = e 0.1t (c).5 = e 15k 15k = ln.5 k = ln.5 k 0.061 15 y = Ce kt (d) ln (0.1) = 0.1t ln (0.1) = t 3 t 0.1 It would take about 3 days. 1 = 100e 0.1t 0.1t When t = 4, y = 9, and when t = 49, y = 15. 4 = Ce 4k 15 = Ce 49k 0.01 = e ln (0.01) = 0.1t ln (0.01) = t 46 t 0.1 It would take about 46 days. 15 Ce 49k 7k = = e

115 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 115 4 Ce 4k 15 7k = ln k = 4 ln ( 15 ) 7 4 0.4 (d) Since the values of k are different, we cannot assume the graph is of the form y = Ce kt. 15. A (t ) = A 0 e kt 1 First, find k. Let A (t ) = A 0 and t = 1.5. 1 1.5k A 0 = A 0 e 1 = e 1.5k 1 ln 1 ln = 1.5k k = 1.5 Now, let A 0 = 1 and let t = 0.5. (Note that the half-life is given in billions of years, so 50 million years is 0.5 billion year.) (continued on next page)

116 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 116 (continued) 1 0. (a) 1 t /13 A(t) = A 0 A (0.5) = 1 e 0.5 ln 1.5 0.87 100/13 Therefore, about 87% of the potassium-40 remains from a creature that died 50 million years ago. 16. Solve for r. Let t = 1997 1980 = 17 1 = 1 (1 + r ) 17 84 50 50 = (1 + r ) 17 84 (b) 1 17 50 r = 1 0.066 = 6.6% 84 No, these numbers do not represent a 4% increase annually. They represent a 6.6% increase. A(100) = 4.0 1 A(100) 0.0193 After 100 years, about 0.0193 gram will remain. 1 t /13 0.1 = 4.0 50 1 17 0.1 = 1 = 1 + r 4.0 84 t 1 0.60 A 0 = A 0 e ( ln / 5600)t 0 ( ln /5600)t 0.60 = e A(100) = 4.0 1 ln ln 0.60 = t t /13 ln 0.05 = ln 13 t = 13 ln 0.05 69.19 ln ( 1 ) It will take about 69 years. t /160 17. A(t ) = A 0 e kt 1. (a) A(t) = A 1 100/160 3.83 5600(ln 0.60) = t ln 417 t 1 18. A 0 = A 0 e 0.053t 5600 The sample was about 4100 years old. (b) After 100 years, about 3.8 grams will remain. 1 0.1 = 4.0 0.1 4 = 1 t /160 t /160 1 = e 0.053t ln 1 = 0.053t ln = 0.053t ln t = 13.1 0.053. (a) t ln 0.05 = ln 1 160 t = 1600 ln 0.05 8515 ln ( 1 ) The half-life is about 8600 years. y = y 0 e kt

117 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 117 The half-life of plutonium 41 is about 13 years. 1 When t = 0, y = 500, so y 0 When t = 3, y = 386. = 500. 19. A 0 = A 0 e 0.00043t 386 500e 3k 386 e 3k = = 1 500 = e 0.00043t e 3k = 0.77 3k = ln 0.77 ln 0.77 ln 1 = 0.00043t ln = 0.00043t ln t = 161 0.00043 The half-life of radium 6 is about 1600 years. k = 0.0863 3 0.0863t y = 500e

118 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 118 ( ) (b) From part (a), we have k = ln 386 500. y = 500e kt = 500e = 500e [ln(386/500) /3]t ln (386/500) (t /3) 3 4. y = 40e 0.004t = 500 [e ln(386/500) ] t / 3 (b) 0 = 0e (a) t = 180 y = 40e 0.004(180) = 40e 0.7 19.5 watts 0.0004t (c) = 500 386 500 1 y 0 = y 0 e 0.0863t ln 1 = 0.0863t ln ( 1 ) t /3 t = 0.0863 8.0 = 500(0.7) t /3 1 = e 1 0.004t ln = 0.004t ln t = 173.9 0.004 It will take about 173 days. (c) The power will never be completely gone. The power will approach 0 watts but will never be exactly 0. 3. (a) The half-life is about 8.0 days. y = y 0 e kt When t = 0, y = 5.0, so y 0 = 5.0 When t = 50, y = 19.5 19.5 = 5.0e 50k 19.5 = e 50k 5.0 50k = ln 19.5 k = 5.0 5. (a) Let t = the number of degrees Celsius. y = y 0 e kt y 0 = 10 when t = 0. To find k, let 11 = 10e 10k e 10k = 11 10 10k = ln 1.1 ln 1.1 k = 0.0095 ln ( 19.5 10 ) 5.0 50 y = 11 when t = 10. 0.00497 The equation is y = 10e 0.0095t. y = 5.0e 0.00497t (b) From part (a), we have k = y = 5.0e kt = 5.0e [(ln 19.5/ 5.0) /50]t (b) Let y = 1.5; solve for t. ln ( 19.5 15 = 10e ) 0.0095t 5.0 50. ln 1.5 = 0.0095t t = ln 1.5 4.7 0.0095 15 grams will dissolve at 4.7 C. = 5.0e ln(19.5/ 5.0) (t /50) = 5. 0 [e ln(19.5/ 5.0) ] t / 50

119 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 119 6. t = 9, T 0 (c) = 5.0(19.5/5.0) t /50 = 5(0.78) t /50 1 y = y e 0.00497t 0 0 1 = e 0.00497t 0.00497t = ln 1 e = 0.05 0.1t = ln 0.05 ln ( 1 ) t = 0.00497 139.47 The half-life is about 139 days. f (t ) = T 7. = 18, C = 5, k = 0.6 + Ce kt 0 f (t ) = 18 + 5e 0.6(9) = 18 + 5e 5.4 18.0 The temperature is about 18.0. f (t) = T 0 + Ce kt 5 = 0 + 100e 0.1t 5 = 100e 0.1t 0.1t t = ln 0.05 30 0.1 It will take about 30 min.

10 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 10 8. C = 14.6, k = 0.6, T 0 = 18, 5π 5π 180 14. = = 100 f (t) = 10 9 9 π f (t) = T 0 + Ce kt 7π 7π 180 10 = 18 + ( 14.6)e 0.6t 15. = 1 1 π 8 = 14.6e 0.6t 0.5479 = e 0.6t 16. ln 0.5479 = 0.6t t = ln 0.5479 1 0.6 180. 90 = 90 π π = 180 4. 135 = 135 π 3π = 180 4 5. 70 = 70 π 3π = 180 6. 30 = 30 π = 16π 180 9 7. 495 = 495 π = 11π 180 4 = 105 5π = 5π 180 = 900 π 17. Let α = the angle with terminal side through ( 3, 4). Then x = 3, y = 4, and It would take about 1 hour for the pizza to thaw. r = x + y = ( 3) + (4) = 5 = 5. sin α = y = 4 cot α = x = 3.4 Trigonometric Functions r 5 y 4 1. 60 = 60 π = π cos α = x = 3 sec α = r = 5 3 r 5 tan α = y = 4 x 3 x 3 csc α = r = 5 y 4 18. Let α = the angle with terminal side through ( 1, 5) Then x = 1, y = 5, and 3. 150 = 150 π 5π = 180 6 r = x + y = 144 + 5 = 169 = 13. sin α = y = 5 cot α = x = 1 r 13 y 5 cos α = x = 1 r 13 tan α = = x 1 sec α = r = 13 x 1 y 5 r 13 csc α = = y 5 19. Let α = the angle with terminal side through (7, 4). Then x = 7, y = 4, and r = x + y = 49 + 576 = 65 = 5. sin α = y = 4 r 5 cot α = x = 7 y 4 8. 510 = 510 π = 17π 180 6 cos α = x = 7 sec α = r = 5

11 Chapter EXPONENTIAL, LOGARITHMIC, Section AND TRIGONOMETRIC.3 APPLICATIONS: FUNCTIONS GROWTH AND DECAY 11 9. r 5 x 7 5π 5π 180 = = 5 tan α = y = 4 csc α = r = 5 4 4 π x 7 y 4 π π 180 0. Let α = the angle with terminal side through 10. = = 10 3 3 (0, 15). Then x = 0, y = 15, and π 11. 13π = 13π 180 r = x + y = 400 + 5 = 65 = 5. = 390 6 6 π sin α = y = 3 cot α = x = 4 r 5 y 3 1. π = π 180 = 45 cos α = x = 4 sec α = r = 5 4 4 π r 5 x 4 tan α = y = 3 cscα = r = 5 8π 8π 180 13. = = 88 x 4 y 3 5 5 π 1. In quadrant I, all six trigonometric functions are positive, so their sign is +.

115 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 115. In quadrant II, x < 0 and y > 0. Furthermore, r > 0. sin θ = y > 0, so the sign is +. r cos θ = x < 0, so the sign is. r tan θ = y < 0, so the sign is. x cot θ = x = 3 y csc θ = r = y 6. When an angle θ of 45 is drawn in standard position, ( x, y) = (1, 1) is one point on its terminal side. Then cot θ = x < 0, so the sign is. y r = 1 + 1 =. sec θ = r < 0, so the sign is. sin θ = y = 1 = x r csc θ = r > 0, so the sign is +. cos θ = x = 1 = y 3. In quadrant III, x < 0 and y < 0. Furthermore, r > 0. sin θ = y < 0, so the sign is. r cos θ = x < 0, so the sign is. r tan θ = y > 0, so the sign is +. x r sec θ = r = x csc θ = r = y 7. When an angle θ of 60 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (1, 3). Then cot θ = x > 0, so the sign is +. r = x + y = 1 + 3 =. y sec θ = r < 0, so the sign is. x csc θ = r < 0, so the sign is. y 4. In quadrant IV, x > 0 and y < 0. Also, r > 0. sin θ = y < 0, so the sign is. r cos θ = x > 0, so the sign is +. r sin θ = y 3 = r x 1 3 cot θ = = = y 3 3 csc θ = r 3 = = y 3 3 8. When an angle θ of 10 is drawn in standard position, ( x, y) = ( 1, point on its terminal side. Then 3) is one tan θ = y < 0, so the sign is. x cot θ = x < 0, so the sign is. y cos θ = r = 1 + 3 =. x 1 = r cot θ = x = 1 = 3

116 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 116 sec θ = r > 0, so the sign is +. x csc θ = r < 0, so the sign is. y sec θ = r = x y 3 3 5. When an angle θ of 30 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 3, 1). Then r = x + y = 3 + 1 =. tan θ = y = 1 = 3 x 3 3 9. When an angle θ of 135 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 1, 1). Then r = x + y = y 1 + 1 =. tan θ = = 1 x cot θ = x = 1 y

117 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 117 30. When an angle θ of 150 is drawn in 36. When an angle of π is drawn in standard 3 standard position, ( x, y) = ( 3, 1) is one position, ( x, y) = (1, 3) is one point on its point on its terminal side. Then terminal side. r = 3 + 1 =. π x 1 3 sin θ = y = 1 cot = = = 3 y 3 3 r cot θ = x = 3 y sec θ = r = = 3 x 3 3 37. When an angle of π 6 is drawn in standard position, one choice of a point on its terminal side is ( x, y) ( 3,1). Then = 31. When an angle θ of 10 is drawn in r = x + y = 3 + 1 =. standard position, one choice of a point on its csc π = r = = terminal side is ( x, y) = ( 3, 1). Then 6 y 1 r = x + y = 3 + 1 =. 38. When an angle of 3π is drawn in standard cos θ = x = 3 position, one choice of a point on its terminal r side is ( x, y) = (0, 1). Then sec θ = r = = 3 r = x + y = 0 + 1 = 1. x 3 3 3π y 1 sin = = = 1 3. When an angle θ of 40 is drawn in r 1 standard position, ( x, y) = ( 1, point on its terminal side. 3) is one 39. When an angle of 3π is drawn in standard position, one choice of a point on its terminal tan θ = y = 3 side is ( x, y) = ( 1, 0). Then x cot θ = x = 1 = y 3 3 3 r = x + y = x 1 = 1. cos 3π = = 1 r π 33. When an angle of 3 is drawn in standard 40. When an angle of π is drawn in standard position, one choice of a point on its terminal position, ( x, y) = ( 1, 0) is one point on its side is ( x, y) = (1, 3). Then terminal side. Then r = 1 + 0 = 1. r = x + y = 1 + 3 =. sec π = r = 1

118 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 118 x sin π = y 3 = 3 r π 34. When an angle of 6 is drawn in standard 41. When an angle of 7π is drawn in standard 4 position, one choice of a point on its terminal side is ( x, y) = (1, 1). Then position, ( x, y) = ( 3, 1) is one point on its terminal side. Then r = x + y = 1 + 1 =. 7π y 1 r = 3 + 1 =. sin = = = 4 r cos π = x 3 = 6 r 4. When an angle of 5π is drawn in standard π position, one choice of a point on its terminal 35. When an angle of 4 is drawn in standard 5π y 1 position, one choice of a point on its terminal side is ( x, y) = (1,1). tan π = y = 1 4 x side is ( x, y) = (0,1). Then tan undefined. = = is x 0

119 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 119 43. When an angle of 5π is drawn in standard 4 position, one choice of a point on its terminal side is ( x, y) = ( 1, 1). Then r = x + y = 1 + 1 =. 50. The sine function is negative in quadrants III and IV. We know that sin (π /6) = 1/. The solution in quadrant III is π + (π /6) = 7π /6. The solution in quadrant IV is π (π /6) = 11π / 6. The two solutions of sin x = 1/ sec 5π = r = = between 0 and π are 7π / 6 and 11π / 6. 4 x 1 44. When an angle of 5π is drawn in standard position, ( x, y) = ( 1, 0) is one point on its terminal side. Then r = cos 5π = x = 1 r 45. When an angle of 1 + 0 = 1. 3π 4 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 1, 1). Then cot 3π = x = 1 = 1 4 y 1 46. When an angle of 5π 6 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 3, 1). Then 51. The tangent function is negative in quadrants II and IV. We know that tan (π / 4) = 1. The solution in quadrant II is π (π / 4) = 3π / 4. The solution in quadrant IV is π (π /4) = 7π /4. The two solutions of tan x = 1 between 0 and π are 3π / 4 and 7π / 4. 5. The tangent function is positive in quadrants I and III. We know that tan (π / 3) = 3, so the solution in quadrant I is π /3. The solution in quadrant III is π + (π /3) = 4π /3. The two solutions of tan x = are π /3 and 4π /3. 3 between 0 and π 53. The secant function is negative in quadrants II and III. We know that sec (π /6) = / 3, so tan 5π = y 1 3 = = the solution in quadrant II is π (π / 6) = 5π / 6. 6 x 3 3 The solution in quadrant III is π + (π / 6) = 7π / 6. The two solutions of is drawn in standard 47. When an angle of 7π 6 position, one choice of a point on its terminal sec x = / 3 between 0 and π are 5π / 6 side is ( x, y) = ( 3,1). Then and 7π / 6. r = x + y = 3 + 1 =. 54. The secant function is positive in quadrants I sin 7π = y = 1 and IV. We know that sec (π / 4) =, so the 6 r solution in quadrant I is π / 4. The solution in quadrant IV is π (π / 4) = 7π / 4. The two 48. When an angle of π 6 is drawn in standard solutions of sec x = between 0 and π are

10 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 10 position, ( x, y) = ( 3, 1) is one point on its π / 4 and 7π / 4. terminal side. Then r = 3 + 1 =. 55. sin 39 0.693 cos π = x = 3 56. cos 67 0.3907 6 r 49. The cosine function is positive in quadrants I 57. tan 13 1.5399 and IV. We know that cos (π /3) = 1/, so the 58. tan 54 1.3764 solution in quadrant I is π /3. The solution in quadrant IV is π (π /3) = 5π /3. The two solutions of cos x = 1/ between 0 and π are π /3 and 5π /3. 59. sin 0.3638 0.3558 60. tan 1.013 1.6004 61. cos 1.353 0.39 6. sin 1.5359 0.9994

11 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 11 63. f ( x) = cos(3x) is of the form f ( x) = a cos(bx) where a = 1 and b = 3. Thus, a = 1 and T = π = π. b 3 64. f ( x ) = 1 sin ( 4π x ) is of the form 70. The graph of y = 1 cos x is similar to the graph of y = cos x except that it has half the 1 f ( x) = a sin(bx) where a = and b = 4π. amplitude and is reflected about the x-axis. Thus, the amplitude is 1 and T = π = π = 1. b 4π π 65. g (t) = sin ( 4 t + ) is of the form 66. s( x) = 3 sin(880π t 7) is of the form that it has times the amplitude, a third of the period, and is shifted 1 unit vertically. s( x) = a sin(bt + c) where a = 3, b = 880π, π Also, π y = cos (3x 4 ) + 1 is shifted 1 units π 71. y = cos (3x 4 ) + 1 has amplitude a =, period T = π = π, phase shift π b 3 g (t ) = a sin (bt + c) where a =, b = 4, c = π /4 = π, and vertical shift d = 1. and c =. Thus, a = = and b 3 1 T = π = π = 8. Thus, the graph of y = cos (3x 4 ) + 1 is b π similar to the graph of 4 π f ( x) = cos x except and c = 7. Thus, a = 3 and T = π = π = 1. b 880π 440 to the right relative to the graph of graph of g ( x) = cos 3x. 67. The graph of y = sin x is similar to the graph of y = sin x except that it has twice the amplitude. 7. y = 4 sin ( 1 x + π ) + has amplitude a = 4, 68. The graph of y = cos x is similar to the π π period T = = = 4π, phase shift b 1

1 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 1 c graph of y = cos x except that it has twice the = π = π b 1, and vertical shift d =. Thus, amplitude. the graph of ( 1 ) y = 4sin x + π + is similar to the graph of f ( x) = sin x except that it has 4 times the amplitude, twice the period, and is shifted units vertically. Also, y = sin ( 1 x + π ) + is shifted π units to the 69. The graph of y = sin x is similar to the graph of y = sin x except that it is reflected about left relative to the graph of ( 1 ) g ( x) = sin x. the x-axis. (continued on next page)

13 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 13 (continued) 73. The graph of y = 1 tan x is similar to the (b) The angle at the lower left is a right angle with measure 90. Since the sum of all three angles is 180, the measures of the remaining two angles sum to 90. Since these two angles are the base angles of an isosceles triangle they are equal, and thus each has measure 45. 77. (a) Since the amplitude is and the period is 0.350, a = and π 1 b π graph of y = tan x except that the y-values of 0.350 = = = b. points on the graph are one-half the y-values b 0.350 π 0.350 of points on the graph of y = tan x. 74. The graph of y = 3 tan x is similar to the graph of y = tan x except that it is reflected about the x-axis and each ordinate value is three times larger in absolute value. Note that (b) = sin π t 0.350 1 = sin π t 0.350 π π t 0.350π = t = = 0.0875 0.350 4π The image reaches its maximum amplitude after 0.0875 seconds. (c) Therefore, the equation is y = sin(π t /0.350), where t is the time in seconds. t = y = sin π () 1.95 π π the points ( 4,3) and ( 4, 3) lie on the 0.350 graph. The position of the object after seconds is 1.95. 78. (a) The period is π = 9.54 π 14.77 There is a lunar cycle every 9.54 days. 75. (a) Since the three angles θ are equal and their sum is 180, each angle θ is 60. (b) The base angle on the left is still 60. The bisector is perpendicular to the base, so the other base angle is 90. The angle formed by bisecting the original vertex angle θ is 30. (b) y = 100 + 1.8 cos (t 6)π reaches a 14.77 maximum value when (t 6)π cos 14.77 = 1 which occurs when t 6 = 0 t = 6 Six days from January 16, 014, is January, 014. (6 6)π

14 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 14 (c) Two sides of the triangle on the left are y = 100 + 1.8 cos 14.77 = 101.8 given in the diagram: the hypotenuse is, and the base is half of the original base of, or 1. The Pythagorean Theorem gives the length of the remaining side (the There is a percent increase of 1.8 percent. (c) On January 31, t = 15. y = 100 + 1.8 cos (1 5 6)π 99.39 vertical bisector) as 1 = 3. 14.77 76. (a) The Pythagorean Theorem gives the length of the hypotenuse as 1 + 1 =. The formula predicts that the number of consultations was 99.39% of the daily mean.

15 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 15 79. P(t) = 7(1 cos π t)(t + 10) + 100e 0.t (a) Since January 1 of the base year corresponds to t = 0, the pollution level P(0) = 7(1 cos 0)(0 + 10) + 100e 0 is = 7(0)(10) + 100 = 100. 4: P.M. (c) For a patient without Alzheimer s, the heights temperature occurs when t 16.37. 16.37 1 = 4.37 and 0.37 hr = 0.37 60 min min. So, t 16.37 corresponds to about (b) Since July 1 of the base year corresponds to t = 0.5, the pollution level is P(0.5) = 7(1 cos π ) (0.5 + 10) + 100e 0.1 = 7() (10.5) + 100e 0.1 58. (c) Since January 1 of the following year corresponds to t = 1, the pollution level is P(1) = 7(1 cos π )(1 + 10) + 100e 0. = 7(0)(11) + 100e 0. 1 (d) Since July 1 of the following year corresponds to t = 1.5, the pollution level is P(1.5) = 7(1 cos 3π ) (1.5 + 10) + 100e 0.3 = 7() (11.5) + 100e 0.3 96. 8. (a) θ = 3, a = 10, B = 170 tan θ = tan 3 (cos10 cot 170 sin 10 ) θ = tan 1 ( tan 3 40 (cos10 cot 170 sin 10 )) (b) θ = 0, a = 10, B = 160 tan θ = tan 0 (cos10 cot 160 sin 10 ) θ = tan 1 ( tan 0 (cos10 cot 160 sin 10 )) 8 80. 83. Solving c 1 = sin θ 1 for c gives c sin θ c = c 1 sin θ. sin θ 1 When θ 1 = 39,θ = 8, and c 1 = 3 10 8, 81. (a) (3 10 8 )(sin 8 ) 8 c =. 10 m/sec. sin 39 84. Solving c 1 = sin θ 1 for c gives c sin θ No, the functions never cross. (b) For a patient without Alzheimer s, the heights temperature occurs when t 14.9. 14.9 1 =.9 and 0.9 hr = 0.9 60 min 55 min. c = c 1 sin θ. sin θ 1 When c 1 = 3 10 8, θ 1 = 46, and θ = 31, 3 10 8 (sin 31 ) c = = 14, 796,150 sin 46.1 10 8 m/sec.

16 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 16 So, t 14.9 corresponds to about :55 P.M. 85. Since the horizontal side of each square represents 30 and the sine wave repeats itself every 8 squares, the period of the sine wave is 8 30 = 40. 86. Since the horizontal side of each square represents 30 and the sine wave repeats itself every 4 squares, the period of the sine wave is 4 30 = 10.

17 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 17 87. (a) 89. T ( x) = 37 sin π (t 101) + 5 365 (a) T (74) = 37 sin π ( 7) + 5 8 F 365 (b) Since the sine function is zero for multiples of π, we can determine the (b) T (11) = 37 sin π (0) 365 + 5 37 F values (s) of t where P = 0 by setting 880π t = nπ, where n is an integer, and solving for t. After some algebraic (c) T (50) = 37 sin π (149) 365 + 5 45 F manipulations, t = n 880 and P = 0 when (d) T (35) = 37 sin π (4) + 5 1 F 365 n =,, 1, 0, 1,,. However, only (e) The maximum and minimum values of values of n = 0, n = 1, or n = produce the sine function are 1 and 1, values of t that lie in the interval [0, 0.003]. Thus, P = 0 when t = 0, respectively. Thus, the maximum value of T is 37(1) + 5 = 6 F and the minimum 1 1 t = 880 0.0011, and t = 440 0.003. These values check with the graph in part (a). (c) The period is T = π = 1. 880π 440 90. (a) Therefore, the frequency is 440 cycles per second. value of T is 37( 1) + 5 = 1 F. (f ) The period is π = 365. π 365 88. T (t) = 60 30 cos t (a) (b) t = 1 represents February, so the maximum afternoon temperature in February is T (1) = 60 30 cos 1 34 F. t = 3 represents April, so the maximum afternoon temperature in April is T (3) = 60 30 cos 3 58 F. (c) (d) (e) Yes; because of the cyclical nature of the days of the year, it is reasonable to assume that the times of the sunset are periodic. (b) The function s(t ), derived by a TI-84 Plus using the sine regression function under the STATCALC menu, is given by s(t ) = 94.087 sin(0.0166t 1.13) + 347.4158. t = 8

18 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 18 represent s Septembe r, so the maximu m afternoon temperat ure in Septembe r is T (8) = 60 30 cos 4 80 F. t = 6 repres ents July, so the maxim um aftern oon temper ature in July is T (6) = 60 30 cos 3 90 F. ture in December is T (11) = 60 30 cos 11 39 F. (c) s(60) = 94.087 sin[0.0166(60) 1.13] + 347.4158 = 36 minutes = 5:6 P.M. s(10) = 94.087 sin[0.0166(10) 1.13] + 347.4158 = 413 minutes + 60 minutes (daylight savings) = 7:53 P.M. s(40) = 94.087 sin[0.0166(40) 1.13] + 347.4158 = 38 minutes + 60 minutes (daylight savings) = 7: P.M. t = 11 rep res ent s De ce mb er, so the ma xi mu m aft ern oo n te mp era

19 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 19 (d) The following graph shows s(t ) and y = 360 (corresponding to a sunset at 6:00 P.M.). These graphs first intersect on day 8. However because of daylight savings time, to find the second value we find where the graphs of s(t ) and y = 360 60 = 300 intersect. These graphs intersect on day 95. Thus, the sun sets at approximately 6:00 P.M. on the 8nd and 95th days of the year. 91. Let h = the height of the building. tan 4.8 = h h = 65 tan 4.8 60. 65 The height of the building is approximately 60. meters. 9. Let x be the distance to the opposite side of the canyon. Then tan 7 = 105 105 x = 06 x tan 7 The distance to the opposite side of the canyon is approximately 06 ft. Chapter Review Exercises 1. False; an exponential function has the form. True 10. True 11. True 1. False; The period of cosine is π. 13. True 14. False. There s no reason to suppose that the Dow is periodic. 15. False; cos(a + b) = cos a cos b sin a sin b 16. True 17. A logarithm is the power to which a base must be raised in order to obtain a given number. It is the inverse of an exponential. We can write the definition mathematically as y = log a x a y = x, for a > 0, a 1, and x > 0. 18. Exponential growth functions grow without bound while limited growth functions reach a maximum size that is limited by some external constraint. 1 19. One degree is 360 of a complete rotation, while one radian is the measure of the central angle in the unit circle that intercepts an arc with length 1. In a circle with radius r, an angle measuring 1 radian intercepts an arc with length r. To convert from degree measure to radian measure, multiply the number of degrees by π. To convert from radian 180 measure to degree measure, multiply the f ( x) = a x. number of radians by 180. π 3. False; the logarithmic function is not defined for a = 1. f ( x) = log a x 0. There s a nice answer to the question of when to use radians vs degrees at http://mathwithbaddrawings.com/013/05/0/ degrees-vs-radians/ 4. False; ln(5 + 7) = ln 1 ln 5 + ln 7 5. False; (ln 3) 4 4 ln 3 since (ln 3) 4 means (ln 3)(ln 3)(ln 3)(ln 3). 6. False; log 10 0 is undefined since 10 x = 0 has no solution. 7. True 1. Let (x, y) is a point on the terminal side of an angle θ in standard position, and let r be the distance from the origin to (x, y). Then y r sin θ = cscθ =, y 0 r y x r cos θ = r tan θ = y, secθ =, x y 0 x 0 cot θ = x, y 0

14 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Section.4 TRIGONOMETRIC FUNCTIONS FUNCTIONS 14 8. False; ln ( ) is undefined. x y 9. False; ln 4 = 0.6667 and ln 8 ln 4 ln 8 = ln(1/ ) 0.6931.. The exact value for the trigonometric functions can be determined for any integer multiple of π or π. 6 4

13 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 13 3. y = ln ( x 9) In order for the logarithm to be defined, 9. 1 x 3 y = 5 x 9 > 0. So, x > 9 x < 3 or x > 3. x 0 1 The domain is (, 3) (3, ). y 15 5 1 5 4. 1 y = e x 1 In order to the fraction to be defined, e x 1 0. So, e x 1 0 e x 1 x 0. The domain is (, 0) (0, ). 5. y = 1 sin x 1 In order to the fraction to be defined, sin x 1 0. So, sin x 1 0 sin x 1. 1 x 1 30. y = x 1 0 1 6. The domain is y 8 4 1 1 π 3π 5π 7π x x,,,,. " y = tan x = sin x cos x In order to the fraction to be defined, cos x 0. The domain is 31. y = log ( x 1) π 3π 5π x x, ±, ±,. y ± " = x 1 x = 1 + y 7. y = 4 x x x 3 5 9 1 0 1 y 0 1 3 y 1 1 16 4 1 4 16 8. y = 4 x + 3 3. y = 1 + log 3 x y 1 = log 3 x 3 y 1 = x x 1 0 1 x 1 1 9 3 1 3 9

14 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 14 y 19 7 4 13 49 4 16 y 1 0 1 3

15 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 15 33. y = ln ( x + 3) 41. e 0.8 =.554 x 1 0 1 The equation in logarithmic form is y 0 0.69 1.1 1.4 1.6 ln.554 = 0.8. 4. 10 1.07918 = 1 The equation in logarithmic form is log 1 = 1.07918. 34. y = ln x x 4 3 1 43. log 3 = 5 The equation in exponential form is 5 = 3. 44. log 3 = 1 9 y 0.8 0. 0.6 The equation in exponential form is 9 1/ = 3. x 1 3 4 45. ln 8.9 = 4.41763 y 0.6 0. 0.8 The equation in exponential form is e 4.41763 = 8.9. 46. log 3.1 = 0.50651 The equation in exponential form is 10 0.50651 = 3.1. Recall that log x means log 10 x. 35. x + = 1 36. x 9 = 3 4 4 1 b = 47. log 3 81 = x 3 x = 81 48. log 3 16 = x 3 x = 16 8 16 x + = 1 4 3 x 3 1 3 x = 3 4 5 x = 4 3 = x = 4 5x = 4 x + = 3 4 4 x = 1 x = 4 x + = 3 5 1 x = 5 37. 9 y + 3 = 7 y (3 ) y + 3 = (3 3 ) y 3 4 y + 6 = 3 3 y x = 1 b 1/ 4 38. = 49. log 4 8 = x 4 x = 8 ( ) x = 3 x = 3 x = 3 50. log 100 1000 = x 100 x = 1000 (10 ) x = 10 3 x = 3 x = 3 4 y + 6 = 3 y 4 4 y = 6

16 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 16 4 1 = b 51. lo 39. 3 5 = 43 The equation in logarithmic form is log 3 43 = 5. 4 1 = 1 = b 16 4 g 5 3k + log 5. 7k 3 = log 5 3k (7 k 3 ) 4 = log 5 (1k ) 5. log 3 y 3 log 3 8 y = log 3 y 3 y = log 3 8 y 4 4 53. 4 log 3 y log 3 x = log 3 y log 3 x 40. 5 1/ = 5 The equation in logarithmic form is log 5 = 1. 5 y 4 = log 3 x

17 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 17 54. 3 log 4 r log 4 r = log 4 (r ) 3 log 4 r 6 61. 1 + m 5 = 15 r 4 = log 4 r = log 4 (r ) 3 5 1/5 1 + m = 15 1/ 5 3 55. 6 p = 17 ln 6 p = ln 17 p ln 6 = ln 17 1 + m = 15 1/ 5 3 56. m 1/ 5 p = ln 17 = 15 1 1.581 3 ln 6 3 z = 11 ln 3 z = ln 11 ( z ) ln 3 = ln 11 ln 11 z = ln 3 57. 1 m = 7 ln 1 m = ln 7 (1 m) ln = ln 7 1 m = ln 7 ln z = ln 11 + 4.183 ln 3 m = ln 7 1 ln m = 1 ln 7 1.807 ln 6. 1 + p 63. 5 m = 3(15 1/ 5 1).156 = 3 1 + p = ± 3 5 5 + p = ±5 3 p = 5 ± 3 log k 64 = 6 p = 5 ± 5 3 p = 5 + 5 3 1.830 or p = 5 5 3 6.830 k 6 = 64 k 6 = 6 k = 58. 1 k = 9 ln 1 k = ln 9 k ln 1 = ln 9 ln 9 k = 0.884 ln 1 64. log 3 ( x + 5) = 5 3 5 = x + 5 43 = x + 5 38 = x x = 119 59. e 5 x 65. log (4 p + 1) + log p = log 3 = 5 ln e 5 x log [ p(4 p + 1)] = log 3 = ln 5 log (4 p + p) = log 3 5 x = ln 5 x = ln 5 + 5 4 p + p = 3

18 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 18 60. x = ln 5 + 5 3.305 e 3x 1 = 14 ln (e 3x 1 ) = ln 14 3x 1 = ln 14 3x = 1 + ln 14 x = 1 + ln 14 1.13 3 4 p + p 3 = 0 (4 p 3)( p + 1) = 0 4 p 3 = 0 or p = 3 p cannot be negative, so p = 3. 4 4 p + 1 = 0 p = 1

19 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 19 66. log (5m ) log (m + 3) = 74. 360 = π log 5m = m + 3 5m = m + 3 75. 405 = 405 π = 9π 180 4 180 5m = 4(m + 3) 76. 5π = 5π = 900 5m = 4m + 1 π m = 14 3π 3π 180 77. = = 135 67. f ( x) = a x ; a > 0, a 1 4 4 π (a) The domain is (, ). 9π 9π 180 78. = = 81 0 0 π (b) The range is (0, ). 3π 3π 180 = = (c) The y-intercept is 1. 79. 10 10 π (d) The x-axis, y = 0, is a horizontal asymptote. 80. 13π 13π 180 = = 117 0 0 π (e) The function is increasing if a > 1. (f) The function is decreasing if 0 < a < 1. 81. 13π 13π 180 = = 156 68. f ( x) = log a x; a > 0, a 1 15 15 π 8. When an angle 60 is drawn in standard (a) The domain is (0, ). (b) The range is (, ). position, one choice of a point on its terminal side is ( x, y) = (1, 3). Then (c) The x-intercept is 1. r = x + y = 1 + 3 =, so (d) The y-axis, x = 0, is a vertical asymptote. sin 60 = y = 3. (e) f is increasing if a > 1. (f) f is decreasing if 0 < a < 1. r 83. When an angle of 10 is drawn in standard position, ( x, y) = ( 1, 3) is one point on its 69. The domain of f ( x ) = a x is the same as the terminal side, so tan 10 = y = 3. range of f ( x ) = log a x and the domain of f ( x ) = log a x is the same as the range of f ( x ) = a x. Both functions are increasing if 84. When an angle of 45 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = (1, 1). Then x

130 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 130 a > 1. Both functions are decreasing if 0 < a < 1. The functions are asymptotic to different axes. r = x + y = 1 + 1 =, so cos( 45 ) = x = 1 =. 70. 90 = 90 π = 90π = π r 180 180 85. When an angle of 150 is drawn in standard position, one choice of a point on its terminal 71. 160 = 160 π = 8π side is ( x, y) = ( 3,1). Then 180 9 r = x + y = 3 + 1 =, so 7. 5 = 5 π = 5π r 3 180 4 sec 150 = = =. x 3 3 73. 70 = 70 π = 3π 180

131 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 131 86. When an angle of 10 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 1, 3). Then r = x + y = 1 + 3 =, so 97. cos 0.815 0.6811 98. cos 0.5934 0.890 99. tan 1.915 3.4868 csc 10 = r = 3 =. 100. The graph of y = cos x appears in Figure 7 y 3 3 in Section 4 of this chapter. To get y = 4 cos x, each value of y in y = cos x must 87. When an angle of 300 is drawn in standard be multiplied by 4. This gives a graph going position, ( x, y) = (1, 3) is one point on its through (0, 4), (π, 4) and (π, 4). terminal side, so cot 300 = x = 1 3 =. y 3 3 88. When an angle of π 6 is drawn in standard position, one choice of a point on its terminal side is ( x, y) = ( 3,1). Then 1 101. The graph of y = tan x is similar to the r = x + y = 3 + 1 =, so sin π = y = 1. graph of y = tan x except that each ordinate 6 r 89. When an angle of 7π is drawn in standard 3 position, one choice of a point on its terminal side is ( x, y) = (1, 3). Then value is multiplied by a factor of 1. Note that the points ( π, 1 ), (0, 0), and ( π, 1 ) lie on 4 4 the graph. r = x + y = 1 + 3 =, so cos 7π = x = 1. 3 r 90. When an angle of 5π is drawn in standard 3 position, one choice of a point on its terminal side is ( x, y) = (1, 3). Then 10. The graph of y = tan x appears in Figure 8 in Section 4 in this chapter. The difference r = x + y = 1 + 3 =, so between the graph of y = tan x and y = tan x is that the y-values of points on the graph of sec 5π = r = =. 3 x 1 91. When an angle of 7π is drawn in standard 3 position, ( x, y) = (1, 3) is one point on its y = tan x are the opposites of the y-values of the corresponding points on the graph of y = tan x. A sample calculation: When x = π, y = tan π = 1.

13 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 13 terminal side. Then r = 1 + 3 =, so 4 4 csc 7π = r 3 = =. 3 y 3 3 9. sin 47 0.7314 93. cos 7 0.3090 94. tan 115.1445 95. sin ( 13 ) 0.8387 96. sin.3581 0.7058

133 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 133 103. The graph of y = sin x is similar to the c c at 3 107. g (t) = + g 0 e graph of y = sin x except that it has two-thirds a a the amplitude and is reflected about the x-axis. (a) If g 0 = 0.08, c = 0.1, and a = 1.3, the function becomes g (t) = 0.1 + 0.08 0.1 e 1.3t. 1.3 1.3 Graph this function on a graphing calculator. 104. y = 17, 000, y 0 = 15, 000, t = 4 (a) y = y 0 e kt 17, 000 = 15, 000e 4k 17 = e 4k 15 ln 1 7 = 4k k = 15 ln 17 [0, 5] by [0.07, 0.09] 15 From the graph, we see that the maximum 0.313 4 value of g for t 0 occurs at t = 0, the (b) So, y = 15, 000e 0.0313t. 45, 000 = 15, 000e 0.0313t 3 = e 0.0313t ln 3 = 0.0313t ln 3 35.1 = k 0.0313 It would take about 35 years. time when the drug is first injected. The maximum amount of glucose in the bloodstream, given by G(0) is 0.08 gram. (b) From the graph, we see that the amount of glucose in the bloodstream decreases from the initial value of 0.08 gram, so it will never increase to 0.1 gram. We can also reach this conclusion by graphing y 1 = G(t ) and y = 0.1 on the same 105. I ( x) 1 I ( x) = 10e 0.3x 10e 0.3x 1 e 0.3x 0.1 0.3x ln 0.1 screen with the window given in (a) and observing that the graphs of y 1 and y do not intersect. x ln 0.1 7.7 0.3 The greatest depth is about 7.7 m. [0, 5] by [0.07, 0.11] 106. Graph y = c(t ) = e t e t on a graphing (c) Using the TRACE function and the graph calculator and locate the maximum point. A calculator shows that the x-coordinate of the in part (a), we see that as t increases, the graph of y 1 = G(t ) becomes almost maximum point is about 0.69, and the y-

134 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 134 coordinate is exactly 0.5. Thus, the maximum concentration of 0.5 occurs at about 0.69 minutes. horizontal, and G(t ) approaches approximately 0.0769. Note that c = 0.1 0.0769. The amount of glucose a 1.3 in the bloodstream after a long time approaches 0.0769 grams. [0, 1.5] by [ 0.1, 0.4]

135 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 135 108. (a) (b) Yes, the data follows a linear trend. The least squares line is log y = 0.7097 log x 0.4480. (c) Solve the equation from part (b) for y. log y = 0.7097 log x 0.4480 log y = log x 0.7097 0.4480 10 log y = 10 0.7097 log x 0.4480 0.7097 y = 10 log x 10 0.4480 y 0.3565x 0.7097 (d) Using a graphing calculator, the 4 π (0.005) 3 = (4.19 10 15 ) t 3 4 π (0.005) 3 3 = t 4.19 10 15 4 π (0.005) 3 ln 3 = t ln 4.19 10 15 4 π (0.005) 3 ln 3 t 4.19 10 = 15 ln 6.8969 It will take about 7 days for the cancer cell to grow to a tumor with a diameter of 1 cm. 0.0+ 0.06 g 0.000165 g 110. m ( g ) = e coefficient of correlation r 0.965. 109. (a) The volume of a sphere is given by V = 4 π r 3. The radius of a cancer cell is 3 (10 ) 5 5 = 10 m. So, [0, 400] by [0, 400] This function has a maximum value of y 345 at x 187.9. This is the largest value for which the formula gives a reasonable answer. The predicted mass of a polar bear with this girth is about 345 kg. 3 3 3 (10 5 111. (a) The first three years of infancy ) V = 4 π r 3 = 4 π 4.19 10 15 cubic meters (b) The formula for the total volume of the cancer cells after t days is V (4.19 10 15 ) t cubic meters. (c) The radius of a tumor with a diameter of 1 cm = 0.01 m is 0.005 m, so V = 4 π r 3 = 4 π (0.005) 3. 3 3 Using the equation found in part (b), we have corresponds to 0 months to 36 months, so the domain is [0, 36]. (b) In both cases, the graph of the quadratic function in the exponent opens upward and the x coordinate of the vertex is greater than 36 ( x 47 for the awake infants and x 40 for the sleeping infants). So the quadratic functions are both decreasing over this time. Therefore, both respiratory rates are decreasing. (c)

136 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 136 11. (a) (d) When x = 1, the waking respiratory rate is y 1 49.3 breaths per minute, and the (b) (c) sleeping respiratory rate is y 38.55. Therefore, for a 1-year-old infant in the 95th percentile, the waking respiratory rate is approximately 49.3 38.55 10.7 breaths per minute higher. S = 1.35 + 104.6 ln A S = 85.49 A 0.3040 (b) Let a = 546, C 0 = 511, C 1 = 634, t 0 = 0.7, t 1 = 6.05, and b = 4. Then, the function is C (t ) = 546 + (634 546) e 3 cos(( π 4)(t 6.05)) 1 + (511 546) e 3 cos(( π 4)(t 0.7 )) 1 = 546 + 88e 3 cos((π 1)(t 6.05)) 1 35e 3 cos((π 1)(t 0.7)) 1 (d) S = 1.35 + 104.6 ln (984.) 74. S = 85.49 (984.) 0.3040 694.7 Neither number is close to the actual number of 41. (e) Answers will vary. 113. P(t) = 90 + 15 sin 144π t The maximum possible value of sin α is 1, while the minimum possible value is 1. Replacing α with 144π t gives 1 sin 144π t 1, 90 + 15( 1) P(t ) 90 + 15(1) 75 P(t ) 105. Therefore, the minimum value of P(t) is 75 and the maximum value of P(t) is 105. (c) C (0.7 ) = 546 + 88e 3 cos((π 1)( 0.7 6.05)) 1 (d) 35e 511 ng dl 3 cos((π 1)( 0.7 0.7)) 1 This is approximately the value of C 0. C (6.05) = 546 + 88e 3 cos((π 1)(6.05 6.05)) 1 35e 634 ng dl 3 cos((π 1)(6.05 0.7)) 1 This is approximately the value of C 1. (e) No, ( ) C t 0 is very close to C 0 because, π 4 114. (a) The period is given by π, so b period = π = 4 hr

13 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 13 when t = t 0, the last term in C 0 and the first term combine to yield C 0. The middle term in C (t 0 ) is small but not 0. It has the value of about 0.3570. Similarly, the first two terms in C (t 1 ) combine to give C 1, while the last term is small, but not 0. 115. (a) The line passes through the points (60, 0.8) and (110,.). m =. 0.8 = 1.4 = 0.08 110 60 50 Using the point (60, 0.8), we have 0.8 = 0.08 (60) + b 0.8 = 1.68 + b 0.88 = b Thus, the equation is g (t ) = 0.08t 0.88.

131 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 131 (b) 117. p = $690, r = 6%, t = 8, m = A = P r 1 + tm m 0.06 8() Per capita grain harvests have been A = 690 1 + 116. (a) increasing, but at a slower rate and are leveling off at about 0.3 ton per person. = 690(1.03) 16 = $11, 075.68 Interest = A P = $11, 075.68 $690 = $4173.68 118. P = $781.36, r = 4.8%, t = 6, m = 4 A = P r 1 + tm m (6)( 4) 0.048 A = 781.36 1 + 4 = 781.36(1.01) 4 = $3703.31 Interest = $3703.31 $781.36 = $91.95 119. P = $1,104, r = 6.%, t = (b) As more terms are added, the function has greater complexity. R 1 (1) = 67.75 + 47.7 cos 1π 1 + 11.79 sin 1π 1 = 67.75 + 47.7 cos π + 11.79 sin π = 67.75 + 47.7 ( 1) + 11.79 (0) 0.0 R (1) = R 1 (1) + 14.9 cos 1π 10. 11. 1. A = Pe rt = 1,104e 0.06() = $13, 701.9 P = $1,104, r = 6.%, t = 4 A = Pe rt A = 1,104e 0.06(4) = 1,104e 0.48 $15, 510.79 A = $1500, r = 0.06, t = 9 A = Pe rt = 1500e 0.06(9) = 1500e 0.54 $574.01 P = $1, 000, r = 0.05, t = 8 6 A = 1, 000e 0.05(8) = 1, 000e 0.40 1.09 sin 1π $17, 901.90 6 13. $1000 deposited at 6% compounded semiannually. = 0.0 + 14.9 cos π 1.09 sin π = 0.0 + 14.9 (1) 1.09 (0) = 34.3 3 R (1) = R (1).86 cos 1π 4 r tm A = P 1 + m To double:

13 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 13 (b) 117. p = $690, r = 6%, t = 8, m = 14.31sin 1π t 0.06 4 (1000) = 1000 1 + = 34.3.86 cos 3π 14.31sin 3π = 34.3.86 ( 1) 14.31(0) 37. R 3 gives the most accurate value. t = 1.03 ln = t ln 1.03 ln t = 1 years ln 1.03 (continued on next page)

133 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 133 (b) 117. p = $690, r = 6%, t = 8, m = (continued) To triple: 0.06 t 3(1000) = 1000 1 + 3 = 1.03 t ln 3 = t ln 1.03 (b) 70, 000 = 100, 000e 0.05t 7 = e 0.05t 10 ln 7 = 0.05t 10 7.1 t It will take about 7.1 years. ln 3 t = 19 years A K ln 1.03 ln 1 + 8.33( ) 16. t = (1.6 10 9 ) ln 14. $100 deposited at 4% compounded quarterly. A = P r tm 1 + m To double: 0.04 t 4 (a) A = 0, K > 0 t = (1.6 10 9 ln[1 + 8.33(0)] ) ln = (1.6 10 9 )(0) = 0 years (100) = 100 1 + 4 (b) t = (1.6 10 9 ln[1 + 8.33(0.1)] ) = 1.01 4t ln ln = 4t ln 1.01 = (1.6 10 9 ln.76596 ) ln ln t = 17.4 = 1, 849, 403,169 4 ln 1.01 or about 1.85 10 Because interest is compounded quarterly, 9 years 15. round the result up to the nearest quarter, which is 17.5 years or 70 quarters. To triple: 3(100) = 100 1 + 0.04 t 4 3 = 1.01 4t ln 3 = 4t ln 1.01 ln 3 t = 4 ln 1.01 7.6 4 Because interest is compounded quarterly, round the result up to the nearest quarter, 7.75 years or 111 quarters. y = y o e kt k (5) (a) 100, 000 = 18, 000e 100 18, 000 = 100, 000e 5k 18 = e 5k 100 (c) As r increases, t increases, but at a slower and slower rate. As r decreases, t decreases at a faster and faster rate 17. (a) Enter the data into a graphing calculator and plot. (b) The sine regression is y = 1.5680 sin (0.54655t.34990) + 50.671 (c) Graph the function with the data. 8 ln = 5k 0.05 k 1 y = 100, 000e 0.05t

134 Chapter EXPONENTIAL, LOGARITHMIC, AND TRIGONOMETRIC Chapter FUNCTIONS REVIEW EXERCISES 134 (b) 117. p = $690, r = 6%, t = 8, m = (d) T = π 11.479 b The period is about 11.5 months.

Chapter EXTENDED APPLICATION 133 Extended Application: Power Functions 1. (a) Yes, the relationship appears to be linear. (b) Y = 5.065 0.389x (c) (c) Species Mass BMR P. boylii 3.0 54.9 P. californicus 47.55 51.97 P. crinitus 15.90 5.1 P. eremicus 1.50 33.11 P. gossypinus 1.50 36.98 P. leucopus 3.00 45.0 P. maniculatus 0.66 39.79 P. megalops 66.0 90.69 P. oreas 4.58 43.51 P. polionotus 1.00 1.48 P. sitkenisis 8.33 46.74 P. truei 33.5 56.70 (d). (a) It is decreasing; the exponent is negative. As the price increases the demand decreases. Y = 158.4 x 0.389 x Y = 3.63x 0.7617 (d) An approximate value for the power is 0.76. (b) Yes, the relationship appears to be linear. Y = 0.7617 x + 1.874

11 Chapter Functions Chapter Functions 11 Chapter Functions a. Calculating Numerical Expressions Spreadsheet can be used as a calculator once we use symbol =. To evaluate numerical expression 7.33 3 14 1 55.1 we type =7.33^3-14*(1+55.1) in any cell. The copy of your expression will appear in formula bar, next to f x. Enter. The result of calculation -545.847 will appear in the cell, as shown in Figure. Figure b. Using Function Notation Excel allows us to create and use our own functions but it also has built-in functions that are performing various operations on data entered in cells. To calculate value f (0.9) for our own function f ( x) 1.8x 5 3x 3 7.3 we enter the value of x in B and we enter the function in cell C as =-1.8*B^5+3*B^3-7.3. Figure 3 shows the final result f (0.9) 6.17588. In case of long algebraic expressions it is easier to type the function inside the formula bar. Notice that instead of x in the formula we are entering the value of cell B. You can type B or obtain the same by clicking on the cell B. Once your function is entered, you can evaluate it for a different value of x, say x.03 by changing cell B value to.03. That way y value will change automatically to -44.553. Figure 3 The list of built-in functions is found in the name box, to the left from the formula bar, after you type symbol = in any cell. Click on arrow next to the name box to obtain the list, like

1 Chapter Functions Chapter Functions 1 in Figure 4. Click on More Functions to search over 100 built in functions sorted in categories like Statistics, Logical etc., like in Figure 5.

13 Chapter Functions Chapter Functions 13 Figure 4 Figure 5 Another way to find and insert a build-in function is by clicking on the tab Formulas and selecting a function from the ribbon, like in Figure 6. If you remember only the beginning of the name of the function, type it in a cell after =. The spreadsheet will help you remember the rest of the name by offering possible choices. Figure 6 The built-in function SUM makes it easy to quickly sum columns, rows, or individual cells of data in a worksheet. To add numbers in cells A1:A4 we ll type = and select built-in function SUM as in Figure 7. One easy way to tell Excel that you want the numbers in cells A1:A4 is by clicking on the small matrix to the right from the Number1 box, and then selecting the array A1:A4. c. Creating Function Tables Figure 7 Excel can graph the function y x 3 5x 1 after the table of (x, y) values of interest is created. Say, we are interested in y values of the function for x 3.0,.5,.0,...,.5, 3.0.

14 Chapter Functions Chapter Functions 14 Enter the first two values -3 and -.5 in cells A and A3. One fast way to type the remaining x values in column A is to select the array A:A3 and drag the handle from the lower right corner down, over column A, until cells A4:A14 are filled with the desired x values, as in Figure 8. Column B is reserved for the y values of our cubic function. A simple way to list the y values under observation is to click on B and enter the formula =A^3-5*A+1. Drag the handle from the lower right corner over column B as in Figure 9 to copy the formula down. Values - 11, -.13, 3, etc. will appear in cells B:B14. Figure 8 Figure 9 d. Graphing Function Once we have table of (x, y) values listed in cells A:B14 we can work on a graph. One of many ways to create a graph of our cubic function y x 3 5x 1 is to first select array A:B14 to indicate x and y coordinates of points under observation. Select the Insert tab and then in the ribbon select the "Scatter" graph to the right. Select the chart sub-type Scatter with smooth lines and markers as shown in Figure 30. Figure 30 The final look of your graph is a matter of your personal taste. Click on your chart and the Chart Tools menu will appear as a new ribbon: use this to upgrade the style. Right-clicking on various sections of the final chart will let you select more options for gridlines, background color, text font, etc. Position your mouse exactly over one of the specially marked data points on your final graph, but do not click. After a short delay, the exact coordinates of that point will appear in a small yellow box next to the point, like in Figure 31.

15 Chapter Functions Chapter Functions 15 Figure 31 e. Piecewise Functions To graph a piecewise defined function we ll create table of (x, y) values we are interested in, insert graph, and then define the domain of each. As an example we ll use the function: x if x f x 6 if x 10 x if x! Figure 3 shows table of values of all three functions: y x, y 6, and y 10 x for x 4.0, 3.5, 3.0,... To graph all three function together on 4, 6 inserted scatter graph with smooth lines. we select A:D and

16 Chapter Functions Chapter Functions 16 Figure 3

17 Chapter Functions Chapter Functions 17 The chart we obtained shows all three functions, each on the entire domain (-4, 4), as shown in Figure 33. We need to adjust the domain of each. Figure 33 Right-click on Series1 in legend under the graph and click on Select Data. The new pop-up screen will help change the visible domain of the first function y x to 4,. Select Series1 and Edit as in Figure 34. Figure 34 Figure 35 The new pop-up screen allowed us to change Series1 name to x< in the legend. Click to encircled box matrix in Figure 35. In the new pop-up screen select the array A:A14 to indicate that the first function domain is 4,. The final graph is shown in Figure 36.

18 Chapter Functions Chapter Functions 18 Figure 36 f. Finding Intersection Points Suppose that the supply function is p 0.q 51 and the demand function is p 3000 / (q 5). We d like to find the exact intersection point of these two functions (the equilibrium point). One way to do it is to look at the table of values and examine y values for supply and demand functions. The first two values of x that are 0 and 5 are entered in cells A and A3, then the array A:A3 is copied down to cells A4:A4 by dragging the fill handle from the lower right corner. The functions are entered in cells B and C as formulas =0.*A+51 and =3000/(A+5), and copied down by dragging the fill handle. Figure 37 shows the equilibrium point in the table. Figure 37 It is even better if we graph the functions together, and find the intersection point on the graph. To obtain the graph select cells array A:B4 and then Scatter with Smooth Lines from the Insert tab ribbon. Position the cursor exactly above the intersection point, but do not click. After a brief delay coordinates of the intersection point appear, like in Figure 38.

19 Chapter Functions Chapter Functions 19. Figure 38 One more way to find the intersection point of functions p 0.q 51 and p 3000 / (q 5) is to find zeros of the function y (0. x 51) 3000 / ( x 5). Type 0 in cell H as shown in Figure 39, and type the formula =(0.*H+51)-3000/(H+5) in cell I. Select What-If Analysis in Data tab ribbon. Click on Goal Seek and fill the obtained pop-up screen as shown in Figure 40. Cell H value will change to 45, that is the first coordinate of the intersection point. Figure 39 Figure 40 g. Finding Maximum and Minimum Points We can estimate the maximum of the function y 0.0037 x 3 0.0591x.534 x 38.1 by creating a spreadsheet table (see chapter c) and examining y values. Once we have the table we might decide to graph the function (see chapter d) and position the mouse over the point. Coordinates (11, 54.008) will appear as in Figure 41.

0 Chapter Functions Chapter Functions 0 Figure 41 By looking at the y values of marked data points from the graph, we can estimate that the percentage will be maximized where the x -value is between 10 and 11. Assume we'd like to know the exact maximum of our function. Spreadsheet programs have a built-in Add-in called Solver. To use it in Excel you need to load it first. Click on File tab, select Options and then select Add-Ins the Microsoft Office Button at the top left corner, and then click Excel Options. Select Add-ins. After you load Solver Add-In to your Add-Ins, the Solver, command is available in Data tab ribbon as one of Analysis tools. Excel s command Solver will calculate y s for different x values in the formula y 0.0037 x 3 0.0591x.534 x 38.1 until the maximum or minimum value for y is found. Assign an initial x-value for the variable cell F as shown in Figure 4. We selected 3, but any value may be used here. Enter the quadratic function spreadsheet formula =-0.0037*F^3-0.0591*F^+.534*F+38.1 in the objective cell G. Select Solver and fill the boxes. Click on Solve and the coordinates of the point where the function is maximized will be determined. In this example, the x-value 10.7 will appear in cell F and the y-value 54.0 will in cell C as the coordinates of the maximum. Figure 4 Calculus for the Life Sciences nd Edition Greenwell SOLUTIONS MANUAL Full download at: https://testbankreal.com/download/calculus-for-the-life-sciences-nd-editiongreenwell-solutions-manual-/ Calculus for the Life Sciences nd Edition Greenwell TEST BANK