Solutions Key Exponential and Logarithmic Functions

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1 CHAPTER 7 Solutions Key Exponential and Logarithmic Functions ARE YOU READY? PAGE D. C. E. A. x ( x ) (x) = x (x) = x 6 6. y -1 (x y ) = (y -1 y ) x = (y)x = 1x y 7. a8 = a (8 - ) a = a 6 9. x y = x ( - 1) y ( - 6) x y 6 = x 1 y -1 = x y 11. (x) ( x ) = 9x ( x ) = 6x 1. a- b = a (- - ) b - (-1) a b -1 = a -6 b = b a 6 1. I = 000(%)() = 180 The simple interest is $ (r)() = r = 90 r = 0.01 or 1.% The interest rate is 1.%. 1. P + P(6%)() = 10 P + P(0.18) = P = 10 P = 00 The loan is $ x - y = x = y + x = y x = y - x = (y - ) x = 6y y 1 y 10 (1-10) = y = y 10. ( x ) - = ( ( x ) -1 ) = ( x) = 7 x 17. y = -7x + y + 7x = 7x = -y + x = -y y = x - 1 y = x - -x = -y - x = y , , EXPONENTIAL FUNCTIONS, GROWTH, AND DECAY, PAGES CHECK IT OUT! 1. growth. P(t) = 0(1.1 ) t The population will reach 0,000 in about 0.9 yr.. v(t) = 1000(0.8 ) t The value will fall below $100 in about 1. yr. 6 Holt McDougal Algebra

2 THINK AND DISCUSS 1. The base of the exponential function is between 0 and 1, so the function shows decay. An exponential decay function decreases over any interval in its domain.. Possible answer: f(x)= 1. 1 x shows growth, and g(x)= 0. 9 x shows decay. The graphs intersect at (0, 1).. Possible answer: exponential decay; exponential growth. 6a. f(x)= (0. ) x b. c. A new softball will rebound less than 1 inch after bounces. PRACTICE AND PROBLEM SOLVING 7. decay 8. growth EXERCISES GUIDED PRACTICE 1. exponential decay. decay. growth 9. growth. decay a. f(x)= 10 ( x ) b. 10a. f(t)= 80(1.0 ) t b. c. The number of ton-miles would have exceeded or would exceed 1 trillion in year, or a. f(x)= 10 (0.9) x c. The number of bacteria after 1 hours will be about 600, Holt McDougal Algebra

3 b. c. About 6 units will remain after 10 minutes. d. It will take about 1.6 min for half of the dose to remain. 1. No; the variable does not contain an exponent. 1. No; 0 x is 0, a constant function. 1. Yes; the variable is in the exponent = 8 (1+.% ) 8 = (1.0 ) 8 = 1,9,9.1 The balance in 008 will be about $1,000,000. t 16. N(t) =.( ) 17. Let x be the number of years needed. 76(1-0% ) x = 0 76 ( 0.7x ) = x x.8 It will take about.8 years for the computer s value to be less than $ ; 0.1; 0.; 1.00;.0;.8; 10.6;.; ; 6.;.0; 1.00; 0.0; 0.16; 0.06; 0.0; a. b. You will owe $119.6 after one year. c. It will take about 18 months for the total amount to reach $100. 1a. 1000(1-0% ) 6 = 1000(0.8 ) 6 16 The rep sold about 16 animals in the 6th month after the peak. b. Let x be the number of month that the rep first sold less than 1000 animals. 1000(1-0% ) x = ( 0.8x ) = x 0.08 x 1 The rep first sold less than 1000 animals in the 1th month. a. A=P ( 1+ r n) nt = 000 ( 1+ % ) = 000(1.01 ) 0 = The investment will be worth $ after years. b. Let x be the number of years needed. 000 ( 1.01x ) = x = x 1 The investment will be worth more than $10,000 after 1 years. c. A=P ( 1+ r n) nt = 000 ( 1+ % 1 ) 1. (0, 1) = 000( ) = 6.60 His investment will be worth $6.60 more after years.. (0, 8,0]. (.868, 100] 6., 768 (00-1) 7a. = 8 = 0.17 = 17% There is a 17% decrease in the amount each day. b. A(t)= 00(1-17% ) t = 00(0.8 ) t c. A(1)= 00(0.8 ) About 6.8 mg will remain after 1 days. 8. N(t)= 6.1(1 + 1.% ) t = 6.1(1.01 ) t t= = 0 N(0)= 6.1(1.01 ) The population will be about 8.1 billion in x ; when x=, they are equal, but x becomes greater quickly as x increases. 0. Possible answer: A company doubles in size each year from an initial size of 1 people; f(p)= 1( ) x ; f()= 1( ) means there are 96 people in yr. TEST PREP 1. B. H. a= 1, b=.. B CHALLENGE AND EXTEND. The degree of a polynomial is the greatest exponent, but exponential functions have variable exponents that may be infinitely large. 6 Holt McDougal Algebra

4 6. x < x <. 7. x > Let x be the cost of a new video game, and y be the cost of an old one. x + y = x + y = 19 Solve equation 1 and equation for x and y using elimination. (x + y) = () 6x + y = 70 -(x + y) = -19 -x - y = -19 x = 7 x = Substitute for x into equation to find y. x + y = 19 () + y = 19 y = 10 y = The cost of a new video game is $, and the cost of an old one is $. 7. odd; positive; 1 8. even; positive; 9. even; negative; 9. ; (, ), (-0.767, 0.88) (0) + = 10 = d + = (10) 1 10 d + = 0 1 d +.6 d 1. There are 100 mosquitoes per acre at the time of the frost; it takes about 1. days for the population to quadruple. 1. If b = 0, f(x) = 0; if b = 1, f(x) = 1. These are constant functions. If b < 0, noninteger exponents are not defined. SPIRAL REVIEW. D: {x x }; use y; f(x) = x shifted right units. D: ; R: {y y 1}; f(x) = x reflected across x-axis and shifted 1 unit up 7- INVERSES OF RELATIONS AND FUNCTIONS, PAGES 98-0 CHECK IT OUT! 1. relation: D: {1 x 6}; R: {0 y } inverse: D: {0 y }; R: {1 x 6} a. f -1 (x) = x b. f -1 (x) = x -. f -1 (x) = x + 7. f -1 (x) = x -. D: ; R: ; f(x) = x stretched vertically by a factor of. D: ; R: ; f(x) = x shifted units down. inverse: z = 6t - 6 = 6(7) - 6 = 6 6 oz of water are needed if 7 teaspoons of tea are used. THINK AND DISCUSS 1. Possible answer: When x and y are interchanged, the inverse function is the same as the original function. The graph of an inverse is the reflection across y = x, but the original function is y = x.. Possible answer: y = x; y = x. Possible answer: You get the original function; yes, the original is a function. 66 Holt McDougal Algebra

5 . EXERCISES GUIDED PRACTICE 1. relation. relation: D: {1 x }; R: {1 y 8}; inverse: D: {1 x 8}; R: {1 y };. relation: D: {-1 x }; R: {- y -1}; inverse: D: {- x -1}; R: {-1 y };. f -1 (x) = x -. f -1 (x) = 1 x 6. f -1 (x) = x 7. f -1 (x) = x f -1 (x) = 1 (x + 1) 9. f -1 (x) = (x - ) 10. f -1 (x) = -x f -1 (x) = - x f -1 (x) = 1 x f -1 (x) = x + PRACTICE AND PROBLEM SOLVING 18. relation: D: {-1 x }; R: {1 y } 19. relation: D: {- x }; R: {- y } inverse: D: {1 y }; R: {-1 x } inverse: D: {- y }; R: {- x } 0. f -1 (x) = 1. 1x 1. f -1 (x) = x + 1. f -1 (x) = 0.x. f -1 (x) = - 1 x + 1. f -1 (x) = 0.08x f -1 (x) = x f -1 (x) = x f -1 (x) = 1.1 x 7. f -1 (x) = -x f -1 (x) = - 1 x f -1 (x) = x f -1 (x) = (x - ) 9. f(x) = 1900x , where x is the number of years after 001; f -1 (x) = x , where x is the number of 1900 bachelor s degrees awarded; f -1 ( ) = About years after 001, 1.7 million bachelor s degrees will be awarded. = - 9 x - x 1 - = - 0a. m = y - y 1 b. The slope of the inverse line is the negative recipical of the slope of the original line, which in this case is - ( 1 -) = F = 9 C + = 9 (16) + = C is about 61 F. 1a. f -1 (x) = 1 - x Holt McDougal Algebra

6 b. f -1 (00) = = 600 The boiling point of water will fall below 00 F above about 600 ft. c. f -1 (160.) = = = 796 The mountain peak s altitude is 7,96 ft.. (, -), (1, ), and (-, -). (, ), (, ), (-, -1), and (-1, -). The inverse is x =, which is a line parallel to the y-axis. So it is not a function.. f(x) = x; f -1 (x) = x = 1.9x f -1 () = 1.9() = 1.8 It will take Warhol 1.8 s to complete a m race. 6a. C = n +. b. n = C -. = = 7 Seven tickets are purchased when the credit card bill is $17.0. c. n =.0 -. = 1.9 No; when C = $.0, n is not an integer. 7. B; the student may have found the inverse of each term. 8. The function is inversed, and the graph is reflected over the line y = x. The result may or may not be a function. 9. Yes; possible answer: for the ordered pairs (, 1) and (, ), the inverse relation is (1, ) and (, ), which is a function. TEST PREP 8. A; f -1 (x) = x + = 1 ( x + ) = 1 x F 0. C 1. x 1 y 0 1 CHALLENGE AND EXTEND (x - b). y = m = x m - b m. ay + bx = c ay = c - bx y = c - bx a. y = - b a x + c a y = x ; switch x and y: x = y. x - y 1 = m(y - x 1 ) x - y 1 = my - m x 1 x - y 1 + m x 1 = my x - y 1 + m x 1 = y m x - y 1 m + x 1 = y 6. Either the function and its inverse are both f(x) = f -1 (x) = x, or the function and its inverse are both f(x) = f -1 (x) = -x + k, where k is any real number constant a. S(c) = 1 c - 7 b. C(s) = s = ( s + 7 ) = s ; yes; head circumference as a function of hat size. 9. c. C ( 7 8) = ( 7 8) = The head circumference of the owner is in. 1. always. sometimes. never. always. always 6. always 7a. P = 17 0 d b. D: {d d 0}; R: {P P 1.7} c. d = P = 0 17 depth as a function of pressure (P - 1.7) = 0 17 P - ; d. At.9 ft, the pressure is.9 psi. SPIRAL REVIEW 60a..9,.18, 6.0, 6.89, 7., 8.16 b. {v.9 v 8.16} 61. (x + )(x - )(x - 1) = 0 6. (x - ) (x + ) = 0 (x + x - 6) (x - 1) = 0 (x - 7x + 6) = 0 x - 1x + 1 = 0 (x -( x - 10 = 0 ) ) = 0 68 Holt McDougal Algebra

7 6. [x - (1 - i)][x - (1 + i)](x - ) = 0 [(x - 1) + i][(x - 1) - i](x - ) = 0 ((x - 1 ) - i ) (x - ) = 0 (x - x (-1)) (x - ) = 0 (x - x + ) (x - ) = 0 (x - x + 6x - ) = 0 x - 8 x + 1x - 8 = 0 6. (x + )(x - 8)(x - 9) = 0 ( x - x - ) (x - 9) = 0 (x - 1 x + 1x + 16) = 0 x - 8 x + x + = 0 6. decay 66. decay 67. growth 68. growth 7- LOGARITHMIC FUNCTIONS, PAGES 0-11 CHECK IT OUT! 1a. lo g 9 81 = b. lo g 7 = c. lo g x 1 = 0 a = 10 b. 1 = 1 c. ( 1 ) - = 8 a. log = - b. lo g 0.0 = -1. D: {x x > 0}; R:. ph = -log ( ) =.8 The ph of the iced tea is.8. THINK AND DISCUSS 1. The inverse of an exponential function is a logarithmic function and vice versa. Exponential functions have a vertical asymptote, and logarithmic functions have a horizontal asymptote. The domain of exponential functions is, and the range is restricted. The domain of logarithmic functions is restricted, and the domain is.. Possible answer: no; lo g 16 =, but lo g 16 = 0... EXERCISES GUIDED PRACTICE 1. x. lo g. 1 = 0. lo g 8 = 1.. log 0.01 = -. lo g = x 6. - = x = = = x 10. lo g 7 = 11. lo g ( 1 9) = - 1. lo g = 1. lo g = 1. f(x): D:, R: {y y > 0}; 1. f(x): D:, R: {y y > 0}; f -1 (x): D: {x x > 0}, R: f -1 (x): D: {x x > 0}, R: y 1 8 f f x poh = -log ( ) = 8. The poh of the water is 8.. PRACTICE AND PROBLEM SOLVING f lo g x =. 18. lo g 6 (16) = x 19. lo g 1. 1 = 0 0. lo g 0. = = 6. 6 = x.. 0 = 1. π 1 = π. lo g 1 = 0 6. log = - 7. lo g 6 = 8. lo g = - 9. f(x): D:, R: {y y > 0}; 0. f(x): D:, R: {y y > 0}; f -1 (x): D: {x x > 0}, R: f -1 (x): D: {x x > 0}, R: y f -1 f x y -1 f -1 x 69 Holt McDougal Algebra

8 1. ph = -log ( ) = 6. No; the ph is 6., so the flowers will be pink. log A - log P log log 1000 a. n = = = 11 log (1.017) log (1.017) The debt has been building for 11 months. log A - log P log 10 - log b. n = = = 9 log (1.017) log (1.017) It will take 9 additional months until the debt exceeds $10. c. As the debt builds, it takes a shorter time for the debt to increase by a relatively constant amount.. 1; lo g a b = 0 means a 0 = b and a 0 is 1 for any a 0. a. Jet: L = 10 log ( 101 I 0 I 0 ) = 10 log ( 10 1) = 10(1) = 10 db; Jackhammer: L = 10 log ( 101 I 0 I 0 ) = 10 log ( 10 1) = 10(1) = 10 db; Hair dryer: L = 10 log ( 107 I 0 I 0 ) = 10(7) = 70 db; Whisper: L = 10 log ( 10 I 0 I 0 ) = 0 db; = 10 log ( 10 7) Leaves rustling: L = 10 log ( 10 I 0 I 0 ) = 10 log ( 10 ) = 10() = 10() = 0 db; Softest audible sound: L = 10 log ( I 0 = 10 log ( 10 ) I 0 ) = 10 log (1) = 10(0) = 0 db b. Let the intensity of the rock concert be x I log ( x I 0 I 0 ) = log (x) = 110 log (x) = 11 x = The intensity of the rock concert is I 0, and it should be placed between jackhammer and hair dryer. c. No; 0 bels is 00 db.. Yes; possible answer: log 1 0 = log 1000 =, and there are zeros after the 1. log = log 0.01 = -, and there are zeros log 00.; log a. orange b. lemon c. grapefruit 8. lo g 0 means 0 x =, and lo g 1 means 1 x =. These have no solution. TEST PREP 9. C 0. G 1. A. F. 6 = 6 lo g 6 = 6 CHALLENGE AND EXTEND. The range of lo g 7 x is negative for 0 < x < 1 and positive for x > 1. The range of lo g 0.7 x is positive for 0 < x < 1 and negative for x > 1.. lo g 9 = ; lo g 7 = ; lo g = lo g 9 + lo g 7 = lo g lo g b ( b x ) + lo g b ( b y ) = lo g b ( b x + y ) 6. Let lo g 7 7 x + 1 = a. Write the above exponental equation as a logarithmic equation, we obtain 7 a = 7 x + 1 a = x + 1 Hence we have proved that lo g 7 7 x + 1 = x a. 11 = 08 Hz, lo g 08 = 11 b. octaves lower; lo g 6 = 8, lo g =, 8 - = SPIRAL REVIEW 8. ( ( a ) ( b )) ( 10a b ) 10 ( a ) ( b ) 100a 8 b 0. -t ( s t -1) (- )s( t t -1 ) -10st 1. 7 a - b (ab + a -1 b ) 9. 8s t 6 s t 8 8s s t6 s 1 t s t 7a - b ab + 7 a - b a -1 b 1a -1 b + 8 a - b. 0 = - 16 t 16t = t = 16 t = 16 t = It took s or 1. s for the brick to hit the ground. t 8 70 Holt McDougal Algebra

9 . 0., 0.9, 1, 1.7, , 1.67, 1, 0.6, ,., 1, 0., PROPERTIES OF LOGARITHMS, PAGES 1-19 CHECK IT OUT! 1a. lo g 6 + lo g lo g (6 ) lo g lo g lo g 1 7 lo g 7 ( 9 7 ) lo g b. lo g lo g () 1 b. lo g lo g 1 9 lo g 1 lo g 1-1 ( 7 1 9) a. log 1 0 log 10 (1) a. 0.9 b. 8x a. lo g 9 7 lo g 7 lo g = log ( E 11.8) ) 10 ( ) 8 = log ( E c. lo g ( 1 ) lo g ( 1 ) (-1) - b. lo g 8 16 lo g 16 lo g = log ( E 11.8) 10 1 = log E - log = log E = log E 10.8 = E = About 6 times as much energy is released by an earthquake with a magnitude of 9. than by one with a magnitude of 8. THINK AND DISCUSS 1. Change the base and enter Y=log(X)/log() is , and is about.98, so is about You get lo g b a = 1 log a b.. EXERCISES GUIDED PRACTICE 1. lo g 0 + lo g 6. lo g (0 6.) lo g 1. lo g + lo g 7 lo g ( 7) lo g 81. log. - log 0.0 log (. 0.0) log lo g 7 9 lo g 7 ( 7 ) lo g x lo g 10 lo g 1. lo g lo g 1 7 lo g lo g 10 log 10 log 1.. log log 1000 log ( ) log lo g 0 - lo g lo g ( 0 ) lo g 6 6. lo g lo g 6. lo g 6 ( ) lo g lo g 1 (0.) ( 0. ) lo g 1 lo g lo g (0. ) lo g ( -1) lo g lo g 8 lo g lo g lo g 7 log 7 log.7 71 Holt McDougal Algebra

10 = log ( E ) ( ) 8.1 = log ( E ) 1.1 = log ( E 11.8) = log E - log = log E = log E 10.9 = E 7.9 = log ( E ) ( ) 7.9 = log ( E ) 11.8 = log ( E 11.8) = log E - log = log E = log E 10.6 = E = About times as much energy was released by the 1811 earthquake than by the 197 one. PRACTICE AND PROBLEM SOLVING 0. lo g 8 + lo g 8 16 lo g 8 ( 16) lo g 8 6. lo g..1 + lo g. lo g. (.1 ) lo g lo g 16 - lo g lo g ( 16 ) lo g 8 6. lo g 1 6 lo g ( ) lo g lo g 1 lo g ( ) 1 lo g log + log log ( ) log log log 100 log ( ) log lo g lo g 1. lo g 1. ( 6.7 ) lo g log (100 ) 0.1 log ( 10 ) 0.1 log x lo g lo g 9 9. lo g 1 16 lo g 16 lo g lo g 9 log 9 log 1.8. lo g 1 lo g 1 lo g. 100 = 10 log ( I I 0 ) 10 = log ( I I 0 ) = I I I 0 = I 10 = 10 log ( I I 0 ) 10. = log ( I I 0 ) = I I I 0 = I I I 0 = I 0.16 The concert sound is bout.16 times more intense than the allowable level. 6a. 1 - (-.) = log d = log d = log d = d d = d The distance of Antares from Earth is about 18 parsecs. b..9 - M = log M = log..9 - M (1.).9 - M = 6.7 M = -.8 The absolute magnitude is about -.9. c. - (-0.) = log d 10. = log d = log d = d d = d The distance to Antares is about 1. times as great as the distance to Rho Oph. 7. lo g b m + lo g b n = lo g b mn 8. lo g b m - lo g b n = lo g b m n 9. lo g b ( bm ) n = lo g b mn nlo g b b m = mn 0. lo g - lo g 18 lo g ( 18) lo g 1-1. log log 1 + log 10 log ( ) log Holt McDougal Algebra

11 . - lo g lo g log lo g (1) lo g 1 + lo g 1 1 lo g 1 ( ) lo g 1 - log log (1) a. log 0 = log ( 10) = log + log b. log 00 = log ( 1 0 ) = log + log c. log 000 = log ( 1 0 ) = log + log a. P = 1(1 - % ) t = 1(0.96 ) t b. t = lo g 0.96 ( P 1) c. log ( x 1) log ) log The population will drop below 0 after about 9 years. d. t = lo g 0.96 ( 0 1) = log ( 0 9. P = 0(1 + 8% ) t = 0(1.08 ) t t = lo g 1.08 ( P 0) = lo g 1.08 ( 0 log ( 0) 0 = 0) log It will take about.9 years for the value of the stock to reach $0. 0a. (00) = 00(1.016 ) n = (1.016) n log = log (1.016) n log = nlog (1.016) log log (1.016) = n.7 n It will take about.7 months for the debt to double. b. It will take another.7 months for the debt to double again. c. no Possible answer: Change the base from b to 10 by writing log b x as log x. Enter log(x)/log(b), using log b the calculator s LOG key. a. lo g = lo g 1 ( 0) = lo g 1 - lo g b. lo g = lo g 1 (0 ) = lo g lo g c. lo g 1 00 = lo g 1 0 = lo g 1 0 (1.). 6a. log b. log ( ) = log. + log c. log (. 1 0 ) = log. + log log ( a 1 0x ) = x + log a d. log ( ) = log. + log Yes, the conjecture holds for scientific notation with negative exponents. 7. sometimes 8. always 9. always 60. never 61. always 6. never 6. sometimes 6. never 6. B; log 80 + log 0 log (80 + 0) TEST PREP 66. B 67. H; lo g 9 x + lo g 9 x lo g 9 x + lo g 9 x lo g 9 x CHALLENGE AND EXTEND 68. A; log 6 log ( ) log + log 69a. Possible answer: the on the top scale is lined up with the 1 on the lower scale. At on the lower scale, the product, 6, is read on the top scale. So the sum of log units and log units is log 6 units. b. The lengths show log + log = log( ) = log {x x < - x > } 71. {x x > 1} 7 Holt McDougal Algebra

12 7. ( x > 1 and x > 0) or 7. {x x > 0} ( x < 1 and x < 0) {x -1 < x < 0 x > 1} 7. {x x > -1} 7. {x -1 x < 0} 76. Let lo g b a p = m b m = a p Let lo g b a = n b n = a Substitute b n for a into equation 1. b m = ( bn ) p b m = b np m = np Substitute lo g b a p for m and lo g b a for n into equation. lo g b a p = plo g b a 77. lo g 9 x lo g 9 ( ) x x lo g 9 9 x(1) x 78. x = x = x = 79. x = x 0 = 1 x = -8 The solution for the x = - equation is Since x > 0, there is no {x x > 0 and x 1}. solution for the equation. SPIRAL REVIEW = (x - 1) 9 = x - -x = -1 x = n = n + 9 7n = 9 n = ( 16) ( -1) ()(i) 1i ( 8) ( -1) ( ) (i) 8i 8. (x + 1) = (x - 6) x + = 6x x = - x = ( n + ) = 10n - 8n + 6 = 10n - -n = -10 n = ( 0) ( -1) - 1 ( 10) (i) -i ( 1) ( -1) ( ) (i) i 89. lo g 1 = 90. log 0.1 = lo g 6 6 = lo g 6 = x = 1 lo g 1 1 = = lo g 16 = = lo g = = 0.0 lo g = -0. READY TO GO ON? PAGE 1 1. decay. growth a. p = 1000(1. ) d. decay. growth b. There will be about 17,086 bacteria in the culture the following Monday f -1 (x) = x f -1 (x) = 1 x f -1 (x) = ( ) - x 11. f -1 (x) = x - 7 Holt McDougal Algebra

13 1. f -1 (x- 10) (x)= f -1 (0.0-10) (0.0) = = 19.0 =. The number of hours of labor is. h. 1. lo g 9 = 1. lo g = 0 1. lo g 0. =- 16. lo g = x 17. = ( 1 ) - = = 1 0. e = x 1.. lo g 81 + lo g 9 lo g (81 9) lo g lo g lo g lo g 1. ( ) lo g lo g 7 lo g lo g 7 lo g lo g 7 0. lo g 6 lo g. lo g 1 lo g 1 lo g lo g 1 ( ) 1. lo g 6 lo g 6 () lo g lo g EXPONENTIAL AND LOGARITHMIC EQUATIONS AND INEQUALITIES, PAGES -8 CHECK IT OUT! 1a. x = 7 x = x= x= 1. c. x = 1 log x = log 1 x log = log 1 log 1 x= log 1.0 b. 7 -x = 1 log 7 -x = log 1 -x log 7 = log 1 -log 1 x= log n log n- 1 log 10 8 (n- 1)log 8 8 n- 1 log n 8 log + 1 n 18 You would receive at least a million dollars on day 18. a. = log 8 + log x = log 8 + log x = log 8 x 10 = 10 log 8 x 1 0 = 8 x 1=x = x a. x= b. x< c. x= 1000 THINK AND DISCUSS b. log x- log = 0 log x= log log x = log x = x= 1. If log a= log b, then 1 0 a = 1 0 b. The exponents must be equal for the exponential expressions to be equal. a. Write log x = 10 as log x. b. Write log x+ log as log x c. Take the log of both sides. d. Use 10 as a base for each side. e. Rewrite log (x+ ) + log (x) as log (x + x). f. Use 6 as a base for each side.. Yes; possible answer: log (-x)= 1 has -10 as a solution. 7 Holt McDougal Algebra

14 . EXERCISES GUIDED PRACTICE 1. exponential equation 1. x = ( ) x = ( ) 1 x = x= x= 8. x = x+ 1 x = ( ) x+ 1 x = x+ x= x+ x=- 6. ( 1 ) x = ( 1 ) x ( ( 1 ) ) x= ( 1 ) x ( 1 ) x = ( 1 ) x x=x x= x+ 1 = 9 log. x+ 1 = log 9 (x+ 1)log. = log 9 x+ 1 = log 9 log. x= ( 1 log 9 log. ) - 1 x x = x- ( ) x = x- x = x- x=x- x=-. x = 10 log x = log 10 x log = 1 x= 1 log = 00( ) t t log.9 log 1.0 t log.9 t log 1.0 log.9 log 1.0 t 6.8 t It will take 7 years for the population to exceed 10,000 people. 9. lo g (7x+ 1) = lo g ( - x) 7x+ 1 = - x 8x= 1 x= lo g 6 (x + ) = 6 log 6 (x+ ) = 6 x+ = 16 x= 1 x= log 7 - log ( x 1. lo g x 9 = 1 9 lo g x= 1 lo g x= x= ) = 0 ) log 7 = log ( x 7 = x 16 = x 108 = x x. 1. lo g 7 ( - x)= lo g 7 ( x ) - x= x = 1x x= log x+ log (x+ 8) = 0 log x(x+ 8) = 0 x(x+ 8) = 10 x + 8x- 100 = 0 (x- )(x+ 0) = 0 x= 16. log ( x+ 10) + log x+ 1 = 0 log x ( x+ 10) =-1 x ( x+ 10) = x + 10 x= x + x- 1 = 0 (x- 1)(x+1)= 0 1. log 0 + log ( x ) = log x= x= 1 0 x= x= 1 76 Holt McDougal Algebra

15 17. x =. 19. x = x 0. x > 10 PRACTICE AND PROBLEM SOLVING 1. x - 1 = 1 6 x - 1 = -6 x - 1 = -6 x = -. ( 1 ) x - x = 1 ( -1) x - = ( ) x - x = x - x = x = x x = 0.8. (1.) x - 1 = 1. log(1.) x - 1 = log 1. (x - 1)log 1. = log 1. log 1. x - 1 = log 1. log 1. x = log x 7.9 x = 1. x log + 1 = log 1. ( x + 1 ) log = log 1. x log = log x = log 1. ( log ) - 1 x.. ( 1 ) x = 8 x - 1 ( -) x = ( ) x - 1 -x = x - -x = x - -x = - x = 0.6. ( 1 ) -x = 1.6 log ( 1 ) -x = log 1.6 -x log 1 = log 1.6 log 1.6 x = - log 1 x > ( 1 ) t > 1 t 1 log 0.1 > log ( 1 ) t 1 log 0.1 > t 1 log 1 log 0.1 log 1 > t 1 1log 0.1 log 1 > t 0.8 > t It takes 1 min for the amount to drop below 0 mg. 8. lo g (7x) = lo g (x + 0.) 7x = x + 0. x = 0. x = lo g ( 1 + x ) = 1 + x x = = 1 x = 0 0. log x - log (1.) = log x 1. x =.1 = 10 x = lo g x =. lo g x =. lo g x = 0.6 x = log x - log ( x 100) = x log ( x x 100) = x log 100 = x = x. - log x = log ( 1) x = log x + log ( 1) x ) = log ( x 10 = x 00 = x 0 = x 77 Holt McDougal Algebra

16 . x = 6. x 1.. x < 1 7. log x = log ( x - 1 ) x = x = x - x = (x - )(x + ) x = or - log(-) is undefined, so the only solution is x =. 8. x = 100 log x = log 100 x log = x log = 1 x = 1 log 1. 0a. = lo g 1 l = 1 l 1 8 = l 9. x + = 6 x + = 6 x + = 6 x = b. 0 = lo g 1 l 0 = 1 l 1 = l n = lo g 1 (1) = lo g 1 = -1 The f-stop setting is f/. n = 0 1 n 1 = 1 n - = 1 - = n 1 - = n The position is keys below concert A.. 00 = 0(1 +.% ) n = (1.0) n log = log (1.0) n log = n log 1.0 log log 1.0 = n 1.7 n It will take at least 16 quarters or yr.. 0; log x = log x, no value of x satisfies the inequality; the graphs coincide, so there is no region where log x < log x.. The student solved log (x + ) = 8.. Method 1: Try to write them so that the bases are all the same. x = 8 x = ( ) x = 9 x = 9 6a. Decreasing; is less than 1. Method : Take the logarithm of both sides. x = 10 log x = log 10 xlog = 1 x = 1 log b. t = = 0 N(0) = 119(0.987 ) 0 = 119(1) = 119 t = = 0 N(0) = 119(0.987 ) 0 9 There are 119,000 farms in 1980 and 9,000 in 000. c = 119(0.987 ) t 67.7 (0.987 ) t 0 t = 010 The number of farms will be about 80,000 in a.. = 18(10 ) h h log h h.9 = 18(10 ) h h log h 11 h The lowest altitude is 11 km, and the highest is km; the model is useful in lower stratosphere and upper troposphere. b. P(0) = 18(10 ) (0) = 18(10 ) 0 = 18(1) = kpa = psi = 18.6 psi The model would predict a sea-level pressure greater than the actual one. TEST PREP 8. B; b x = c log b x = log c x log b = log c x = log c log b 0. B 9. J; log (x - 1) = - log x log (x - 1) + log x = log x(x - 1) = x(x - 1) = 1 0 x - 1x = 0 (x - )(x + ) = 0 x = 78 Holt McDougal Algebra

17 CHALLENGE AND EXTEND 1. Possible answer: no; x = x x, and x 1 = x x, so x = 1, but lo g 1 is not defined.. x = 0.1 log x = ( -) log x = -lo g x = log - x = - x = 1 or lo g 6 - lo g x > 1 lo g ( 6 x ) > 1 6 > 1 x {x x < 1} x < 1 SPIRAL REVIEW a. 0.7x + 0.y. b. He can buy six -by-6-inch photographs.. deta = (7) - (1) = 6 6. deta = -1(10) - 9(-) = 7. deta = 1 (6) - 0(-1) = 8. deta = (9) - 1 (6) = 9. f -1 (x) = x f -1 (x) = 1 6 x + = 1 x f -1 (x) = (x - 9) = x f -1 (x) = x + 1 = 7 7 x THE NATURAL BASE, e, PAGES 1-6 CHECK IT OUT! 1. a. ln e. =. b. e ln x c. ln e x + y = x + y ln x e x. A = Pe rt = 100 e 0.0(8) 1.1 The total amount is $ = 1 e-k(8) ln 1 = ln e-8k ln -1 = -8k -ln = -8k k = ln = 60 e -0.08t = e-0.08t ln 00 = ln e-0.08t 60 ln = -0.08t t = ln I will take about 7.6 days to decay. THINK AND DISCUSS 1. Possible answer: e and π are irrational constants. π is a ratio of parts of a circle and is greater than e.. e x and ln x are inverse functions. ln represents the logarithm, or exponent, when e is used as a base.. 79 Holt McDougal Algebra

18 EXERCISES GUIDED PRACTICE 1. f(x) = ln x; natural logarithm ln e 0 = ln e a = a 19. e ln (c + ) = c + 0. e ln x = e ln x = x 1. A = 000 e 0.0(). The total amount of his investment after years is about $... 1 = 1 e-k(110) ln 1 = ln e-110k.. ln -1 = -110k -ln = -110k k = ln N(000) = 0 e (000) 17 1 = 0 e t 1 0 = e t ln 1 = ln e t 0 6. ln e 1 = 1 7. ln e x - y = x - y 8. ln e ( -x ) = - x 10. e ln x = e ln x = x 9. e ln x = x 11. A = Pe rt = 770 e 0.0() The total amount is $ = 1 e-k(6) ln 1 = ln e-6k ln -1 = -6k -ln = -6k k = ln N() = 0 e -0.11() 16 The amount remaining after hours is about 16 mg. PRACTICE AND PROBLEM SOLVING ln 1 0 = t t = ln The decay constant is ; the amount remaining is 17 g after 000 years; it takes 100,000 years for the 0 grams to decay to 1 gram. a. ln 10.0; log e 0.; they are reciprocals. b. ln 10 = log10 loge = 1 log e. log x = ln x by change of base, so ln 10 ln 10 (log x) = ln 10 ( ln x ln 10) = ln x a. 10 = (06-70) e -0.8t 10 = 16 e -0.8t = e-0.8t ln = -0.8t t = ln It takes about. min for the coffee to reach its best temperature. b. 10 = (06-86) e -0.8t 10 = 10 e -0.8t = e-0.8t ln = -0.8t t = ln It takes about.8 min for the coffee to reach its best temperature. 80 Holt McDougal Algebra

19 c. room: 17. min; patio: 16.9 min 6. Possible answer: They are the same. y = lo g 6 x is changed to base e by y = ln x and to base 10 by ln 6 y = log x log B 8. A 9. C 0a. t = = = 700 e 0k = e0k ln 7 7 = 0k ln 7 k = The growth factor k is about 0.1. b. t = = 6 P(6) = 700 e 0.1(6) The population would have been about.6 million in ln + ln x = 1 ln x = 1 x = e 1 x = e 0.. ln 10 + ln x = 10 ln 10 x = x = e 10 x = e10 10 x = ± e x = ± e 10 ±7. ln - ln x = ln x = x = e x = e 0.. ln x - = 0 ln x = ln x = 1 x = e. ln x - ln x = 0 ln x - ln x = 0 0 = 0 {x x > 0} 7a. b. : y = 0 and y = 1 6. e ln x = 8 ln e ln x = ln 8 ln x = ln 8 x = 8 x = c. Possible answer: The epidemic spreads slowly at first, then steadily, and then it tapers off slowly at the end. 8a. 1: f(x) = 1 0 x ; : f(x) = e x ; : f(x) = x ; b. (0, 1) c. 0 = = e 0 = 1 9. Possible answer: a little more; $1000 at 8% interest compounded daily for 1 year: A = 1000 ( ) 6 $108.8; compounded continuously: A = 1000 e 0.08 $ a. t = = = 00 e 10k = e10k ln 08 = 10k k = ln b. t = = 0 N(0) = 00 e (0) 8000 There will be about 8,000 farms in 010. c. 179 = 109 e 10k = e10k ln = 10k k = ln A(t) = A 0 e kt = 109 e 0.006(0) 10 The average size will be about 10 acres in 010. TEST PREP 1. C. J. A. Possible answer: ln ( 1 x) 81 Holt McDougal Algebra

20 CHALLENGE AND EXTEND. Let n be the number of periods needed in one year. ( n )n e 0.08 Solve the equation by trial and error to get n =. Possible answer: ; yes, at 18% interest it takes 17 periods. 6. D: ; R: 0 < y 7a. f(x) = ln (-x) b. f(x) = -ln x c. f(x) = -ln (-x) d. 1 π one asymptote: x = 0. lo g 8 + lo g 1 lo g 8 1 lo g. lo g - lo g 187 lo g 187 lo g ( 1 9) - 7. lo g lo g lo g lo g lo g 6 - lo g 1 lo g 6 1 lo g 6 6. lo g + lo g 1 lo g 1 lo g 8. log x - log x log x x log x 7-7 TRANSFORMING EXPONENTIAL AND LOGARITHMIC FUNCTIONS, PAGES 7- CHECK IT OUT! 1. x j(x) SPIRAL REVIEW 8a. g(x) = 1 f(x) c. g(x) = f(x) b. g(x) = f(x + ) a.. y = 0; translation units right 1 ; y = 0; f(x) = x ; vertical compression by a factor of 1. b. ; y = 0; j(x) = x ; reflection across y-axis and vertical stretch by a factor of 9. g(x) = f(x) + = -x + x g(x) = f(x + ) = -x - x g(x) = -f(x) = x - x +. g(x) = f ( x ) = - 1 x + x - x = -1; the graph of f(x) = ln x is translated 1 unit left, reflected across the x-axis, and translated units down.. g(x) = log (x + ) 8 Holt McDougal Algebra

21 . 0 > -1 log (t + 1) > -1 log (t + 1) 17 < log (t + 1) < t + 1 t > The average score will drop to 0 after 8,680 years, and it is not a reasonable answer. THINK AND DISCUSS 1. Possible answer: {x x < 0}. Possible answer: The transformations of x and f(x) are the same for the functions f(x) + k, f(x - h), af(x), f ( 1 b x ), -f(x), and f(-x).. Possible answer: f(x) = a x : vertical translations and reflections across the x-axis change the range; f(x) = lo g b x: horizontal translations and reflections across the y-axis change the domain; no.. EXERCISES GUIDED PRACTICE 1.. x g(x) y = ; translation units down; R: {y y > } x h(x) y = -; translation units down; R: {y y > -} x j(x) y = 0; translation 1 unit left ; y = 0; vertical stretch by a factor of - 1 ; y = 0; vertical compression by a factor of 1 and reflection across the x-axis; R:{y y < 0} -1; y = 0; reflection across both axes; R: {y y < 0] ; y = 0; vertical compression by a factor of 1 -; y = 0; vertical stretch by a factor of and reflection across the x-axis; R: {y y < 0} 9. 1; y = 0; horizontal compression by a factor of 1 8 Holt McDougal Algebra

22 x= 0; vertical stretch by a factor of. 1. g(x)=-0.7 ( x + ) g(x)= 1 log (x- 1) < 6 + ln (t+ 1) 11 < ln (t+ 1) 11 < ln (t+ 1) 11 e < t > 8.1 x= 0; translation units left and vertical stretch by a factor of.; D: {x x>-} x= 0; reflection across the y-axis, vertical compression by a factor of 1, and translation 1. units up t > e The model is translated 1 unit left, stretched by a factor of, and translated 6 units up; the height will exceed 17 feet after about 9 years. PRACTICE AND PROBLEM SOLVING x g(x) y=-1; translation 1 unit down; R: {y y>-1} x h(x) y= 0; translation units left 18. x j(x) y=-1; translation 1 unit right and 1 unit down; R: {y y>-1} ; y= 0; vertical stretch by a factor of -0.; y= 0; vertical compression by a factor of 0. and reflection across the x-axis; R: {y y< 0} ; y= 0; vertical stretch by a factor of and reflection across the y-axis ; y= 0; vertical compression by a factor of 0. -1; y= 0; horizontal stretch by a factor of and reflection across the x-axis; R: {y y< 0} -; y= 0; vertical stretch by a factor of and reflection across both axes; R: {y y< 0} 8 Holt McDougal Algebra

23 7. x = ; translation units right; D: {x x > } x = 0; vertical stretch by a factor of and reflection in the x-axis. 8. f(x) = -1. ( 1 ) x - 9. f(x) = ln (x + ) f(x) = e (1 - x ) t e - 17 t e t ln t -17ln x = -; translation units left; vertical compression by a factor of, and translation units down; D: {x x > -} x = 0; vertical stretch by a factor of and reflection across the x-axis The model is horizontally stretched by a factor of 17, reflected across the y-axis, and vertically stretched by a factor of 870. The instruments will function properly for about 7 years. Possible answer: -t Use P(t) 900() 100 to approximate the function. Round the predicted value of t to 0 yr. -0 P(0) 900() The estimate confirms that after 7 yr, the power output will be about 600 W.. vertical stretch by e ; g(x) = e e x. A. E. D 6. C 7. F 8. The graph of y = x- + is a translation units right and units up of the graph of y = x. The asymptote is y =, and the graph approaches this line as the value of x decreases. The domain is still all real numbers, the range changes to values greater than, and the intercept is 0- + =.. D: ; R: {y y > }; x-intercept: none; y-intercept: (o,.) 9. The graph of 0. never y = log(x + ) is a vertical stretch by a factor of and a translation units left of the graph of y = logx. The vertical asymptote changes to x = -. The domain changes to numbers greater than - and the range is still all real numbers, and the intercept is log(0 + ).9. D: {x x > -}; R: ; x-intercept: -; y-intercept: (o,.9) 1. always. never. sometimes a. (1000) = 1000 ( 1 + r ) () = ( 1 + r ) 0 ln = 0ln ( 1 + ) r ln 0 = ln ( 1 + ) r ln e 0 = 1 + r ln r = ( e 0-1) 0.11 A rate of about 1.1% will double the investment in years. b. (1000) = 1000 ( = t ln = tln t = ln ln It will take about 0 years to double the investment at a rate of.%. c. A(10) = 1000 ( ) t ) (10) The total amount will be about $ after 10 years. 8 Holt McDougal Algebra

24 . C 6. A 7. B 8a. vertical stretch by a factor of b. horizontal translation units right; D: {x x > } c. horizontal stretch by a factor of d. Possible answer: horizontal compression since 0.9 t = 0.97 kt log 0.9, where k = the constant log 0.97, which is > 1 9. Changing h translates the graph right (+) or left (-), and changing k translates the graph up (+) or down (-). 0. Possible answer: Translation left or right: Replace x with x - h. Translation up or down: Add k. Reflect across x-axis: Multiply b x by -1. Reflect across y-axis: Replace x with -x. Vertical stretch or compression: Multiply b x by a ±1. Horizontal stretch or compression: Divide x by c ±1. 1a. N(t) = 1 (17)(0.99)t = 19(0.99 ) t m b. N(m) = 19(0.99 ) 1 c. m = 1 + = N(17) = 19(0.99 ) 1 1 There were about 1 soybean farms at the end of May TEST PREP. A. H. B CHALLENGE AND EXTEND a. translation 1 unit up b. horizontal compression by 1 10 c. They are equivalent. d. log (10x) = log 10 + log x = 1 + log x 6. The value of y is undefined for x -. ed: s/b a?? 7. For (x - h) > 1, f(x) > 0. For (x - h) = 1, f(x) = 0. For 0 < (x - h) < 1, f(x) < 0. For (x - h) < 0, f(x) < 0 is undefined. SPIRAL REVIEW 8. f(x) = x - x + =( x - x ) + =( x - x ) = (x - 1 ) + min: (1, ); D: ; R: {y y > }; 9. f(x) = x + x - = ( x + 1 x ) - = ( x + 1 x + 1 6) - - ( 1 6) 16 = ( x + 1 8) - 81 min: ( - 1 8, ) ; D: ; R: {y y > }; 60. f(x) = - x - x + 1 = - ( x + x) + 1 = - ( x + x + 1 ) = - ( x + 1 ) + max: ( - 1, ) ; D: ; R: {y y < }; 61. f(x) 0.0 x x x ln e x + = x + 6. ln e -x = -x 6. e ln (x - 1) = x e ln x = x 7-8 CURVE FITTING WITH EXPONENTIAL AND LOGARITHMIC MODELS, PAGES -1 CHECK IT OUT! 1a. yes; 1.. no. B(t) 199(1. ) t ; (1. ) t t 199 ln 000 tln ln 000 t 199 ln ed: s/b a? The number of bacteria will reach 000 in about 10. min. 86 Holt McDougal Algebra

25 . S(t) ln t; ln t ln t ln t 7.1 t e The speed will reach 8.0 m/s in about 16.6 min. THINK AND DISCUSS 1. if there is a common ratio between the data values. Possible answer: There is only 1 ratio between data values, so there is no way to determine if that ratio is constant.. EXERCISES GUIDED PRACTICE 1. exponential regression. no. yes;. no. yes; 6. T(t) 11.(0.9 ) t ; Enter the data and use an exponential regression. Graph the function and find where the value falls below 0. It will take about 1.6 min for the tea to reach the temperature. 7. P(t) ln t; ln t ln t ln t 778. t e 11 1 It will take about 1 mo for the population to reach PRACTICE AND PROBLEM SOLVING 8. no 9. no 10. yes; yes; 1 1. P(t) a 9.(1.0 ) t ; 10 < 1.6(1.0 ) t 10 < 1.0t 1.6 ln 10 < tln ln 10 t > 1.6 > 7.8 ln = 07 The population will exceed 10 million in T(t).(1.17 ) t ; 100 <.(1.17 ) t 100 < 1.17t. ln 100. < tln1.17 ln 100. t> > 19.8 ln = 011 The number of telecommuters will exceed 100 million in t(p) ln p; t(00) ln = 00 The population will reach 00 in year yes; f(x)= 1.(7. ) x 16. yes; f(x)= (0. ) x 17. Possible answer: a third data point because points can be fit by many different functions 18. r(d) 10.99(0.999 ) d ; r(000) 10.99(0.999 ) The calf survival rate is 1.0 per 100 cows at snow depths of 000 mm. 19. Use year 00 as the starting year (t= 0). s(t) 68.(.6878 ) t ; t= + = s() 68.(.6878 ) 6,,00 The sales will be about 6,,00 in three years. 0. Possible answer: an exponential decay model; the ratios are nearly constant. 1. Possible answer: nth differences have the same common ratio as first differences. a. F(t)= 011.6(0.98 ) t b = = 1.6% c. t= = 0 F(0)= 011.6(0.98 ) 0 10 There will be about 1,0,000 acres of farmland in Holt McDougal Algebra

26 a. t = 1; S = 100(0.79 ) S = 79. S = 0. t = ; S = 100(0.79 ) S = 6.0 S = 6.8 t = S = 100(0.79 ) S = 16.0 S = 8.0 The speed is 0. mi/h at 1 s, 6.8 mi/h at s, and 8.0 mi/h at 8 s. b. s = 100 ( 0.8t ) ; the value of a is accurate to the nearest whole number; the value of b is accurate to the nearest thousandth.. Possible answer: There will be a constant ratio between consecutive values rather than constant differences. a. exponential b. Linear; the log of an exponential function of x is linear. TEST PREP 6. C 7. F 8.. (. ) = 6.1 CHALLENGE AND EXTEND 9. 8 = a b Solve the system for a and b. 00 = a b Substitute for a: 00 = ( 8 b ) b. Solve for b: 6. = b b =. (b > 0). Solve for a: a = So f(x) = 7.68(. ) x. 0a. f(t) 0.01 ( 0.96t ) b. The initial concentration was 0.01 mg/c m, and it was not above the health risk level. c > 0.01 ( 0.96t ) > 0.96t 0.01 ln > t ln ln t > 0.01 ln 0.96 > 7.7 This will be about 7 hours later. 1. Exponential: {y y 0} causes an error. Logarithmic: {x x 0} or {y y 0} causes an error. SPIRAL REVIEW. -x = -x = ± x = ±9. x - = x - = ± x = ± x = 7 or 1 6. x + x - = 0 (x - 1)(x + ) = 0 x - 1 = 0 or x + = 0 x = 1 or x = - 8. x + 10x + 1 = 0 (x + x + 6) = 0 (x + )(x + ) = 0 x + = 0 or x + = 0 0. x = - or x = = x + -6 = ( ) x + -6 = (x + ) - = x + x = -8 x. 8 = ( 1 ) x + ( ) x = ( -1) x + x -(x + ) = x = -x - x = - x = x = 1 7 x = - x = -. x + = 0 x + = 0 x = -. x + 1 = 10 x = 9 x = ±9 x = ± 9 7. x + x = 0 x(x + 8) = 0 x = 0 or x + 8 = 0 x = 0 or x = x + 9x - 6 = 0 (x - )(x + 1) = 0 x - = 0 or x + 1 = 0 x = or x = x = x + ( ) x = x + x = x + x = x =. 1 6 x = 6 x ( 6 ) x = 6 x x = x x = 0 READY TO GO ON? PAGE x. 9 x + < 7 x. 1 x - 1 = 91 (x - 1)ln 1 = ln 91 x - 1 = ln 91 ln 1. lo g (x - 1) x - 1 x 6 ln 91 ln x = x 0.9 ( 7 ) x + < 7 (x + ) < x x < -8 x < x + = 0 (x + )ln = ln 0 x + = ln 0 ln x = ln 0 ln - x lo g x = 1 x = 1 ( x ) = ( ) x = 1 =, Holt McDougal Algebra

27 7. log 16x - log = log 16x = log x = x = 1 0 x = 9. A = P(1 + r) n A = (1 + r)n P log A = nlog (1 + r) P log A P n = log (1 + r) log n ( ) log ( ) log x + log (x + ) = 1 log x(x + ) = 1 x(x + ) = x + x - 10 = 0 (x - )(x + ) = 0 x - = 0 or x + = 0 x = or x = - x = since x must be positive. It will take about 0. quarters, or yrs, for the account to contain at least $ x 1. ln e = 1. ln e = x 16. e ln (1 - a) = 1- a 17. ln e b + = b + 18a = e-70k ln 1 = -70k k = ln b. N 1000 = 10 e (1000) 8.87 There will be about 8.87 g left after 1000 years. 1.; y = 0; The graph is stretched vertically by a factor of ; y = 0; The graph is stretched horizontally by a factor of. 1. 0; x = -1; The graph is shifted 1 unit left and stretched vertically by a factor of.; D: {x x > - 1}. f(x) = -0. x. yes; constant ratio: ; f(x) = (1.) x ; x = -; The graph is shifted units left and reflected across the x-axis; D: {x x > -}. linear function; f(x) = 0.9x STUDY GUIDE: REVIEW, PAGES natural logarithmic function. asymptote. inverse relation LESSON 7-1. growth 6. decay 8. growth 9. P(t) = 76(1.0 ) t 10.. growth 7. growth yr 89 Holt McDougal Algebra

28 LESSON P T = P L (1-0.0) 1. P L = P T K = 8 M; mi = 8 () km = 0 km LESSON lo g = 18. lo g 9 1 = lo g 1 7 = - 0. = = = x y D: {x x > 0}; R: LESSON 7-9. lo g 8 + lo g 16 lo g (8 16) lo g lo g 18 - lo g lo g 18 lo g 6 6. lo g lo g ( ) lo g 0. log log 10,000 log (100 10,000) log 1,000, log 10 - log 0.1 log log 100. log log 1 0 log ( ) 9 log L = 10 log I I 0 L 10 = log I I 0 L = I I 0 L I = I 0 L + 10 = 10log I I 0 L + 10 = log I 10 I 0 L = I I 0 L + 10 I = I 0 L + 10 L L L I I 0 = = 10 The sound today was 10 times more intense than the sound yesterday. LESSON 7-6. x - 1 = 1 9 x - 1 = - x - 1 = - x = log x >..log x >. log x > 1 x > ( 1 ) x 6 ( -1) x 6 -x 6 x = 0( ) n = 1.0 n ln = n ln 1.0 n = ln ln It will take about years to double the money. LESSON 7-6 0a. t = = 6 19 = e 6k 19 = e6k ln 19 = 6k ln 19 k = b. t = = 80 P(80) = e 0.06(80) 9 The population will be about 9 in 00. LESSON g(x) = -e x - 90 Holt McDougal Algebra

29 . y-intercept: 0.6; y = 0; vertically compressed by a factor of and horizontally compressed by a factor of 1 6. V (t) = 00(1-0. ) t.. vertically stretched by a factor of 00 x-intercept: 0.; x = -0.; translated 1 unit left and vertically stretched by a factor of 6. f -1 (x) = x f -1 (x) = 6 ( x) 7. f -1 (x) = 6 (x ) 9. f -1 (x) = 6 ( x) LESSON f(x) 11.6(1.0 ) x 7. f(x) ln x 8. the exponential function; r 0.9 versus r 0.60 for the logarithmic function CHAPTER TEST, PAGE 8 1. decay. decay 10. lo g 16 = lo g 16 1 = ( 1 ) - = = 1 1. f -1 (x) = lo g 1 x 1. f -1 (x) = lo g. x. growth. growth D: {x x > 0}; R: D: {x x > 0}; R: 16. f -1 (x) = -lo g x. f(x) = 100(1-0.1 ) t The value will fall below $000 in the 10 th year. D: {x x > 0}; R: 17. lo g 18 - lo g 8 lo g 18 8 lo g lo g lo g ( ) 10 lo g 10 x 1. x - 1 = 79 x - 1 = ( 6) x x - 1 = 6 ( x ) x - 1 = x x = lo g lo g lo g (1.8 ) lo g lo g x = x x 1. - x 1. - x x Holt McDougal Algebra

30 . lo g (x + 8) = x+ 8 = x= 16. log 6 x - log x= 1 log ( 6x x )= 1 log x= 1 x= 10 1 x= 1. > 1(0.9 ) x 1 > 0.9x ln 1 > xln ln 1 x > > 1. ln 0.9 It will take about 1. min to reduce to less than ml. 1 = e-000k ln 1 =-000k k= ln e () There will be about g left after years. 7. f(x)= ln (x+ ) f(x)= ln x; 100 < ln x 1.6 <.6 ln x < ln x 1.6 x>e.6 > 8.1 The population will exceed 100 in year 8. 9 Holt McDougal Algebra