Chapter 3 Exponential, Logistic, and Logarithmic Functions

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1 Chapter Exponential, Logistic, and Logarithmic Functions Section. Exponential and Logistic Functions 75 Section. Exponential and Logistic Functions Exploration. The point (0, ) is common to all four graphs, and all four functions can be described as follows: Domain: (- q, q) Range: (0, q) Always increasing Not symmetric Bounded below by y = 0, which is also the only asymptote lim f(x) = q, lim x: q f(x) = 0 Bounded below by y = 0, which is also the only asymptote lim g(x) = 0, lim g(x) = q x: q [, ] by [, 6] y = a b x y = a b x [, ] by [, 6] y = x [, ] by [, 6] y = a 4 b x [, ] by [, 6] y = x [, ] by [, 6] [, ] by [, 6] y =4 x [, ] by [, 6] y 4 = a 5 b x Exploration. fx = x y 4 =5 x [, ] by [, 6]. The point (0, ) is common to all four graphs, and all four functions can be described as follows: Domain: (- q, q) Range: (0, q) Always decreasing Not symmetric. [ 4, 4] by [, 8] fx = x gx = e 0.4x [ 4, 4] by [, 8] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

2 76 Chapter Exponential, Logistic, and Logarithmic Functions fx = x gx = e 0.5x 5. Translate f(x) = x by units to the right. Alternatively, g(x) = x- = - # x = 8 # x = 8 # f(x), so it can be obtained from f(x) using a vertical shrink by a factor of. 8 [ 4, 4] by [, 8] fx = x gx = e 0.6x [, 7] by [, 8] 7. Reflect f(x) = 4 x over the y-axis. [ 4, 4] by [, 8] fx = x gx = e 0.7x [ 4, 4] by [, 8] [, ] by [, 9] 9. Vertically stretch f(x) = 0.5 x by a factor of and then shift 4 units up. fx = x gx = e 0.8x k = 0.7 most closely matches the graph of f(x).. k L 0.69 Quick Review.. -6 = -6 since (-6) = / =( ) / = = a 6 [ 4, 4] by [, 8] 9..4, since (.4) 5 = Section. Exercises. Not an exponential function because the base is variable and the exponent is constant. It is a power function.. Exponential function, with an initial value of and base of Not an exponential function because the base is variable. 7. f(0) = # 5 0 = # = 9. fa b = - # / = -. f(x) = # a x b. f(x) = # () x = # x/ [ 5, 5] by [, 8]. Reflect f(x) = e x across the y-axis and horizontally shrink by a factor of. [, ] by [, 5]. Reflect f(x) = e x across the y-axis, horizontally shrink by a factor of, translate unit to the right, and vertically stretch by a factor of. [, ] by [, 4] 5. Graph (a) is the only graph shaped and positioned like the graph of y = b x, b Graph (c) is the reflection of y = x across the x-axis. 9. Graph (b) is the graph of y = -x translated down units.. Exponential decay; lim f(x) = 0; lim f(x) = q x: q. Exponential decay: lim f(x) = 0; lim f(x) = q x: q Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

3 5. x Section. Exponential and Logistic Functions 77 [, ] by [ 0., ] 7. x 6 0 [ 0.5, 0.5] by [0.75,.5] 49. [, ] by [, 9] Domain: (- q, q) Range: (0, q) Always increasing Not symmetric Bounded below by y = 0, which is the only asymptote lim f(x) = q, lim f(x) = 0 x: q 9. y = y, since x + 4 = (x + ) = ( ) x + = 9 x y-intercept: (0, 4). Horizontal asymptotes: y=0, y =. 4. y-intercept: (0, 4). Horizontal asymptotes: y = 0, y = [ 0, 0] by [ 5, 5] [ 5, 0] by [ 5, 0] [, ] by [, 8] Domain: (- q, q) Range: (0, q) Always increasing Not symmetric Bounded below by y = 0, which is also the only asymptote lim f(x) = q, lim f(x) = 0 x: q Domain: (- q, q) Range: (0, 5) Always increasing Symmetric about (0.69,.5) Bounded below by y = 0 and above by y = 5; both are asymptotes lim f(x) = 5, lim f(x) = 0 x: q For #5 5, refer to Example 7 on page 85 in the text. 5. Let P(t) be Austin s population t years after 990. Then with exponential growth, P(t) = P 0 b t where P 0 = 465,6. From Table.7, P(0) = 465,6 b 0 = 656,56. So, b = [, 4] by [, 7] 656,56 B 0 465,6 L.050. Solving graphically, we find that the curve y = 465,6(.050) t intersects the line y=800,000 at t L Austin s population passed 800,000 in Using the results from Exercises 5 and 5, we represent Austin s population as y=465,6(.050) t and Columbus s population as y=6,90(.08) t. Solving graphically, we find that the curves intersect at t L.54. The two populations were equal, at 74,86, in Solving graphically, we find that the curve.79 y = intersects the line y=0 when +.40e x t L Ohio s population stood at 0 million in (a) When t = 0, B = 00. (b) When t = 6, B L False. If a>0 and 0<b<, or if a<0 and b>, then f(x) = a # b x is decreasing. 6. Only 8 x has the form a # b x with a nonzero and b positive but not equal to. The answer is E. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

4 78 Chapter Exponential, Logistic, and Logarithmic Functions 6. The growth factor of f(x) = a # b x is the base b. The 88 answer is A. 7. b=.0 6B (a) 9 9. b= 0.6 4B 67 (b) Domain: (- q, q) Range: c e, q b Intercept: (0, 0) Decreasing on (- q, -]. Increasing on [-, q) Bounded below by y = - e Local minimum at a -, - e b Asymptote y = 0 lim [ 5, 5] by [, 5] f(x) = q, lim x: -q f(x) = 0 [, ] by [ 7, 5] Domain: (- q, 0) h (0, q) Range: (- q, -e] h (0, q) No intercepts Increasing on (- q, -]; Decreasing on [-, 0) h (0, q) Not bounded Local maxima at (-, -e) Asymptotes: x=0, y = 0 g(x) = -q lim g(x) = 0, lim x: -q 67. (a) y f(x) decreases less rapidly as x decreases. (b) y as x increases, g(x) decreases ever more rapidly. 69. a Z 0, c =. 7. a 7 0 and b 7, or a 6 0 and 0 6 b Since 0<b<, lim ( + a # b x ) = q and x: -q lim ( + a # c b x ) =. Thus, lim and x: - q + a # b x = 0 c lim. + a # b x = c Section. Exponential and Logistic Modeling Quick Review (.07)() 5. b = 60 = 4, so b = ;4 = ;. 40 Section. Exercises For # 0, use the model P(t) = P 0 ( + r) t.. r=0.09, so P(t) is an exponential growth function of 9%.. r= 0.0, so f(x) is an exponential decay function of.%. 5. r=, so g(t) is an exponential growth function of 00%. 7. f(x) = 5 # ( + 0.7) x = 5 #.7 x (x = years) 9. f(x) = 6 # ( - 0.5) x = 6 # 0.5 x (x = months). f(x) = 8,900 # ( ) x = 8,900 # x (x=years). f(x) = 8 # ( ) x = 8 #.05 x (x = weeks) 5. f(x)= 0.6 # x/ (x=days) 7. f(x)= 59 # - x/6 (x=years) 9. f 0 =.,.875 =.5 = r +, so. f(x)=. #.5 x (Growth Model). For #, use f(x) = f 0 # b x. f Since f(5) = 4 # b 5 0 = 4, so f(x) = 4 # b x. = 8.05, bfi= b= L.5. f(x) L 4 #.5. 5B x 4, 4 c For # 8, use the model f(x) =. + a # b x 40. c=40, a=, so f()= = 0, b = 40, + b 40 60b=0, b=, thus f(x)=. + # a x b 8 5. c=8, a=7, so f(5)=, + 7b = =+4bfi, 4bfi=96, bfi=, b=, thus f(x) L. 5B 4 L # x 0 7. c=0, a=, so f()= + b = 0, 0 = 0 + 0b, 0b =0, b = b=,, B L thus f(x)=. + # 0.58 x 9. P(t) = 76,000(.049) t ; P(t)=,000,000 when t 0.7 years, or the year 00.. The model is P(t) = 650(.075) t. (a) In 95: about P(5),5. In 940: about P(50) 4,65. (b) P(t)=50,000 when t years after 890 in (a) y = 6.6a t/4, where t is time in days. b Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

5 Section. Logarithmic Functions and Their Graphs 79 (b) After 8. days. 5. One possible answer: Exponential and linear functions are similar in that they are always increasing or always decreasing. However, the two functions vary in how quickly they increase or decrease. While a linear function will increase or decrease at a steady rate over a given interval, the rate at which exponential functions increase or decrease over a given interval will vary. 7. One possible answer: From the graph we see that the doubling time for this model is 4 years. This is the time required to double from 50,000 to 00,000, from 00,000 to 00,000, or from any population size to twice that size. Regardless of the population size, it takes 4 years for it to double. 9. When t=, B 00 the population doubles every hour. For #4 4, use the formula P(h)=4.7 # 0.5 h/.6, where h is miles above sea level. 4. P(0)=4.7 # 0.5 0/.6 =.4 lb/in 4. The exponential regression model is P(t) = (.0) t, where P(t) is measured in thousands of people and t is years since 900. The predicted population for Los Angeles for 007 is P(07) L 478, or 4,78,000 people. This is an overestimate of 44,000 people, 44,000 an error of L 0.09 = 9%.,84,000 The equations in #45 46 can be solved either algebraically or graphically; the latter approach is generally faster. 45. (a) P(0)=6 students. (b) P(t)=00 when t.97 about 4 days. (c) P(t)=00 when t 6.90 about 7 days. 47. The logistic regression model is P(x) =, where x is the e x number of years since 900 and P(x) is measured in millions of people. In the year 00, x = 0, so the model predicts a population of P(0) = e (0) = e = L 0.6 L 0,600,000 people [0, 00] by [0, 0] 5. False. This is true for logistic growth, not for exponential growth. 5. The base is.049=+0.049, so the constant percentage growth rate is 0.049=4.9%. The answer is C. 55. The growth can be modeled as P(t)= # t/4. Solve P(t)=000 to find t The answer is D (a) P(x) L where x is the number of e -0.07x, years since 900 and P is measured in millions. P(00) 77.9, or 77,900,000 people. (b) The logistic model underestimates the 000 population by about.5 million, an error of around.%. (c) The logistic model predicted a value closer to the actual value than the exponential model, perhaps indicating a better fit. 59. sinh(-x) = e-x - e -(-x) = e-x - e x = -a ex - e -x b = -sinh(x), so the function is odd. e x - e -x sinh(x) 6. (a) cosh (x) = e x + e -x e x - e -x = # e x - e -x = e x -x = tanh(x). e x + e -x + e e -x - e -(-x) (b) tanh( x)= e -x + e = e-x - e x -(-x) e -x + e x = - ex - e -x = -tanh(x), so the function is odd. e x + e -x e x - e -x (c) f(x)=+tanh(x)=+ e x + e -x e x + e -x + e x - e -x e x = e x + e -x = e x + e -x e x = e x a + e -x e -x b = + e -x, which is a logistic function of c=, a=, and k=. Section. Logarithmic Functions and Their Graphs [ 0,0] by [0,400] P(x) L where x is the number of e x years after 800 and P is measured in millions. Our model is the same as the model in Exercise 56 of Section.. Exploration. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

6 80 Chapter Exponential, Logistic, and Logarithmic Functions 4. Starting from y=ln x: translate left units.. Same graph as part. Quick Review.. 5 = = = 5 = 7. 5 / 9. [ 6, 6] by [ 4, 4] a / e b = e -/ [ 5, 5] by [, ] 4. Starting from y=ln x: reflect across the y-axis and translate up units. [ 4, ] by [, 5] 45. Starting from y=ln x: reflect across the y-axis and translate right units. Section. Exercises. log 4 4= because 4 =4. log =5 because 5 = 5. log 5 5 = because 5 / = 5 7. log 0 = 9. log 00,000=log 0 5 =5. log =log 0 / 0 =. ln e = 5. ln =ln e = e 7. ln 4 e=ln e /4 = 4 9., because b logb = for any b>0. 0 log (0.5) = 0 log 0 (0.5) = 0.5 e log e 6. e ln 6 = =6 5. log and log ( 4) is undefined because 4<0. 9. ln and e ln ( 0.49) is undefined because 0.49<0.. x=0 =00 5. x=0 = = f(x) is undefined for x>. The answer is (d). 9. f(x) is undefined for x<. The answer is (a). [ 7, ] by [, ] 47. Starting from y=log x: translate down unit. [ 5, 5] by [, ] 49. Starting from y=log x: reflect across both axes and vertically stretch by. [ 8, ] by [, ] 5. Starting from y=log x: reflect across the y-axis, translate right units, vertically stretch by, translate down unit. [ 5, 5] by [ 4, ] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

7 Section.4 Properties of Logarithmic Functions 8 5. year and y is the population. Graph the function and use TRACE to find that x L 0 when y L 50,000,000. The population of San Antonio will reach,500,000 people in the year 0. [, 9] by [, ] Domain: (, q) Range: ( q, q) Always increasing Not symmetric Not bounded Asymptote at x= lim f(x)=q Domain: (, q) Range: ( q, q) Always decreasing Not symmetric Not bounded Asymptotes: x= lim [, 8] by [, ] f(x) = -q [970, 00] by [600,000,,700,000] 6. True, by the definition of a logarithmic function. 65. log The answer is C. 67. The graph of f(x)=ln x lies entirely to the right of the origin. The answer is B. 69. f(x) ` x ` log x Domain ` ( q, q) ` (0, q) Range ` (0, q) ` ( q, q) Intercepts ` (0, ) ` (, 0) Asymptotes ` y=0 ` x=0 [, 7] by [, ] Domain: (0, q) Range: ( q, q) Increasing on its domain No symmetry Not bounded Asymptote at x=0 lim f(x) = q 59. (a) ı= 0 log a 0- b = 0 log 0 = 0() =0 db - 0 (b) ı= 0 log a 0-5 =70 db 0 - b = 0 log 07 = 0(7) (c) ı= 0 log a 0 =50 db 0 - b = 0 log 05 = 0(5) 6. The logarithmic regression model is y = ln x, where x is the [ 6, 6] by [ 4, 4] e 7. b= e. The point that is common to both graphs is (e, e). [.7, 5.7] by [., 4.] 7. Reflect across the x-axis. Section.4 Properties of Logarithmic Functions Exploration. log ( # 4) , log +log Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

8 8 Chapter Exponential, Logistic, and Logarithmic Functions. log a , log 8-log b 7. log x=log x / =log x ln x+ ln y=ln x +ln y =ln (x y ). log , log (0.00) log (xy)- log (yz)=log (x 4 y 4 )-log (y z ) 4. log 5=log a 0 =log 0-log b =log a x4 y 4 log a x4 y y z b = z b = In # 8, natural logarithms are shown, but common (base-0) 5. log 6=log 4 =4 log.04 logarithms would produce the same results. log =log 5 =5 log.5055 log 64=log 6 =6 log.8068 ln log 5=log 5 = log 5= log a 0 ln L.8074 b ln L.487 =(log 0-log ).9794 ln 8 log 40 = log (4 # 0) = log 4 + log 0 L.6006 ln log 50 = log a = -ln ln 0.5 ln L b = log 00 - log L ln x 9. log x= The list consists of,, 4, 5, 8, 6, 0, 5,, 40, 50, 64, ln and 80. ln(a + b). log (a+b)= ln Exploration log x. False. log x= log. False; log (7x)=log 7+log x log(x + y) log(x + y). True 5. log / (x+y)= = - log (/) log 4. True 7. Let x=log x b R and y=log b S. 5. False; log =log x-log 4 Then b x =R and b y =S, so that 4 R 6. True S = bx b y = bx - y 7. False; log 5 x =log 5 x+log 5 x= log 5 x log. 8. True b a R S b = log b b x - y = x - y = log b R - log b S Quick Review.4. log 0 =. ln e = x 5 y - 5. x 5 y ( 4) =x y x y -4 = x 7. (x 6 y ) / =(x 6 ) / (y ) / = y (u v -4 ) / - u v 9. = (7 u 6 v - 6 / ) u v - = u Section.4 Exercises. ln 8x=ln 8+ln x= ln +ln x. log =log -log x x 5. log y 5 =5 log y 7. log x y =log x +log y = log x+ log y x 9. ln ln x -ln y = lnx- ln y y = x. log = (log x-log y)= 4 log x - 4B y 4 4 log y. log x+log y=log xy 5. ln y-ln =ln(y/) 9. Starting from g(x)=ln x: vertically shrink by a factor /ln [, 0] by [, ] 4. Starting from g(x)=ln x: reflect across the x-axis, then vertically shrink by a factor /ln 0.9. [, 0] by [, ] 4. (b): [ 5, 5] by [, ], with Xscl= and Yscl= (graph y=ln(-x)/ ln 4). 45. (d): [, 8] by [, ], with Xscl= and Yscl= (graph y=ln(x-)/ ln 0.5). Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

9 Section.4 Properties of Logarithmic Functions [, 9] by [, 7] Domain: (0, q) Range: ( q, q) Always increasing Asymptote: x=0 lim f(x)=q ln (8x) f(x)=log (8x)= ln () 49. [0, ] by [0, 5] (d) ln(y)=5.00 ln x+.0 (e) a 5, b so f(x)=e x 5 =ex 5.7x 5. The two equations are the same. 65. (a) log(w) p log(r) Domain: ( q, 0)ª (0, q) Range: ( q, q) Discontinuous at x=0 Decreasing on interval ( q, 0); increasing on interval (0, q) Asymptote: x=0 lim [ 0, 0] by [, ] f(x)=q, lim x: q f(x)=q, 5. In each case, take the exponent of 0, add, and multiply the result by 0. (a) 0 (b) 0 (c) 60 (d) 80 (e) 00 (f) 0 ( = 0 0 ) I 5. log 0.005(40)= 0.094, so = I= # L lumens. 55. From the change-of-base formula, we know that f(x) = log x = ln x 0.90 ln x. ln = # ln x L ln f(x) can be obtained from g(x) = ln x by vertically stretching by a factor of approximately True. This is the product rule for logarithms. 59. log =log ( # 4)=log +log 4 by the product rule. The answer is B. 6. ln x 5 =5 ln x by the power rule. The answer is A. 6. (a) f(x)=.75 # x 5.0 (b) f(7.) 49,66 (c) ln(x) p ln(y) [, ] by [.6,.8] (b) log r=( 0.0) log w+.6 (c) [, ] by [.6,.8] (d) log r=( 0.0) log (450)+.6.58, r 7.69, very close. (e) One possible answer: Consider the power function y=a # x b then: log y=log (a # x b ) =log a+log x b =log a+b log x =b(log x)+log a. which is clearly a linear function of the form f(t)=mt+c where m=b, c=log a, f(t)=log y and t=log x. As a result, there is a linear relationship between log y and log x. For #67 68, solve graphically x (a) [, 9] by [, 8] Domain of f and g: (, q) Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

10 84 Chapter Exponential, Logistic, and Logarithmic Functions (b) [0, 0] by [, 8] Domain of f and g: (5, q) (c) [ 7, ] by [ 5, 5] Domain of f: ( q, ) ª (, q) Domain of g: (, q) Answers will vary. log x 7. Let y=. By the change-of-base formula, ln x y= log x log x = log x # log e = log e L 0.4. log x log e Thus, y is a constant function. Section.5 Equation Solving and Modeling Exploration. log (4 # 0) L.6006 log (4 # 0 ) L.6006 log (4 # 0 ) L.6006 log (4 # 0 4 ) L log (4 # 0 5 ) L log (4 # 0 6 ) L log (4 # 0 7 ) L log (4 # 0 8 ) L log (4 # 0 9 ) L log (4 # 0 0 ) L The integers increase by for every increase in a power of 0.. The decimal parts are exactly equal # 0 0 is nine orders of magnitude greater than 4 # 0. Quick Review.5 In # 4, graphical support (i.e., graphing both functions on a square window) is also useful.. f(g(x)) = e ln(x/) = e ln x = x and g(f(x)) =ln(e x ) / = ln(e x ) = x.. f(g(x)) = ln(ex ) = (x) = x and (/ ln x) ln g(f(x))= e = e x = x * 0 8 km 7. 60,000,000,000,000,000,000, (.86 * 0 5 )(. * 0 7 ) = (.86)(.) * = * 0 Section.5 Exercises For # 8, take a logarithm of both sides of the equation, when appropriate.. 6 a x/5 b = 4 a x/5 b = 9 a x/5 b = a b x 5 = x = 0. # 5 x/4 = 50 5 x/4 = 5 5 x/4 = 5 x 4 = x = x/ = 0, so -x/ =, and therefore x = x = 0 4 = 0, x - 5 = 4 -, so x = = 5.5. ln 4.. x = ln.06 = log L 4.5. e 0.05x = 4, so 0.05x = ln 4, and therefore x = ln 4 L e -x =, so -x = ln, and therefore x = -ln L ln(x-)= and therefore, so x - = e/, x = + e / L We must have x(x + ) 7 0, so x 6 - or x 7 0. Domain: (- q, -) h (0, q) ; graph (e). x. We must have 7 0, so x 6 - or x 7 0. x + Domain: (- q, -) h (0, q) ; graph (d).. We must have x 7 0. Domain: (0, q); graph (a). For #5 8, algebraic solutions are shown (and are generally the only way to get exact answers). In many cases solving graphically would be faster; graphical support is also useful. 5. Write both sides as powers of 0, leaving 0 log x = 0 6, or x =,000,000. Then x = 000 or x = Write both sides as powers of 0, leaving 0 log x4 = 0, or x 4 = 00. Then x = 0, and x = ;0. 9. Multiply both sides by # = # x, or ( x ) - # x, leaving ( x ) - x - = 0. This is quadratic in x, leading to x ; = = 6 ; 7. Only Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

11 Section.5 Equation Solving and Modeling is positive, so the only answer is 49. Substituting known information into T(t)= T leaves T(t)=+70e ln(6 + 7) kt m + (T 0 - T m )e -kt. x = = log ln (6 + 7) L Using T()=50=+70e k, we have e k = 5, so. Multiply both sides by e x, leaving (e x ) + = 8e x, or (e x ) - 8e x + = 0. This is quadratic in e x, leading to k= - Solving T(t)=0 yields ln 5 L e x = 8 ; 64-4 t 8.4 minutes. = 4 ; 5. Then 5. (a) x=ln(4 ; 5) L ; and therefore 00 = + 5e0.x, so e 0.x = 50 = 0.06, x = ln 0.06 L Multiply by, then combine the logarithms to obtain [0, 40] by [0, 80] x + ln so x + = x. x = 0. Then x + x = e 0 (b) =, The solutions to this quadratic equation are x = ; + 7. ln[(x-)(x+4)]= ln, so (x-)(x+4)=8, or x + x - 0 = 0. This factors to (x-4)(x+5)=0, so x=4 (an actual solution) or x= 5 (extraneous, since x- and x+4 must be positive). 9. A $00 bill has the value of 000, or 0, dimes so they differ by an order of magnitude of =.5. They differ by an order of magnitude of Given b = 0 log I I 0 = 95 b = 0 log I = 65, I 0 we seek the logarithm of the ratio I /I. I 0 log - 0 log I = b I 0 I - b 0 0 alog I I 0 - log I I 0 b = log I I = 0 I = ; L.08. log = I The two intensities differ by orders of magnitude. 45. Assuming that T and B are the same for the two quakes, we have 7.9 = log a - log T + B and 6.6= log a - log T + B, so =. =log(a /a ). Then a /a = 0., so a L 9.95a the Mexico City amplitude was about 0 times greater. 47. (a) Carbonated water: log [H + ]=.9 log [H + ]=.9 [H + ]=0.9.6*0 4 Household ammonia: log [H + ]=.9 log [H + ]=.9 [H + ]=0.9.6*0 [H + ] of carbonated water (b) [H + ] of household ammonia = 0-.9 = (c) They differ by an order of magnitude of 8. [0, 40] by [0, 80] T(x) L # 0.9 x (c) T(0)+0= C 5. (a) [0, 0] by [0, 5] (b) The scatter plot is better because it accurately represents the times between the measurements. The equal spacing on the bar graph suggests that the measurements were taken at equally spaced intervals, which distorts our perceptions of how the consumption has changed over time. 55. Logarithmic seems best the scatterplot of (x, y) looks most logarithmic. (The data can be modeled by y=+ ln x.) [0, 5] by [0, 7] 57. Exponential the scatterplot of (x, y) is exactly exponential. (The data can be modeled by y= # x.) [0, 5] by [0, 0] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

12 ` 86 Chapter Exponential, Logistic, and Logarithmic Functions 59. False. The order of magnitude of a positive number is its common logarithm. 6. x = x = 5 x-=5 x= The answer is B. 6. Given R = log a T + B = 8. R = log a, T + B = 6. we seek the ratio of amplitudes (severities) a /a. alog a T + Bb - alog a T + Bb = R - R a log T - log a = T a log = a a = 0 =00 a The answer is E A logistic regression af(x) = most e b x closely matches the data, and would provide a natural cap to the population growth at approx..0 million people. (Note: x=number of years since 900.) 67. (a) [0, 0] by [ 5, ] 7. One possible answer: We map our data so that all points (x, y) are plotted as (ln x, y). If these new points are linear and thus can be represented by some standard linear regression y=ax+b we make the same substitution (x : ln x) and find y=a ln x+b, a logarithmic regression. 7. x.066 [, 5] by [, 6] x 6.75 (approx.) [, ] by [, 8] As k increases, the bell curve stretches vertically. Its height increases and the slope of the curve seems to steepen. (b) As c increases, the bell curve compresses horizontally. Its slope seems to steepen, increasing more rapidly to (0, ) and decreasing more rapidly from (0, ). 69. (a) r cannot be negative since it is a distance. (b) [, ] by [0, 0] [, ] by [0, ] [ 0, 0] by [ 0, 0] [0, 0] by [-5, ] is a good choice. The maximum energy, approximately.807, occurs when r L log x - log 7 0, so log(x/9) 7 0. Then x =, so x 7 9. Section.6 Mathematics of Finance Exploration. k ` A 0 ` ` ` ` ` ` ` ` 05. ` A approaches a limit of about y=000e is an upper bound (and asymptote) for A(x). A(x) approaches, but never equals, this bound. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

13 Quick Review # 0.05 = 7. # 7.5% =.85% = 0.65 = 65% x=48 gives x= (+0.05)=5 dollars Section.6 Mathematics of Finance 87 In #9 0, the time must be rounded up to the end of the next compounding period ln 9. Solve a + = : t= 4 4 ln.0475 L.4 round to years months. For # 4, use the formula S=Pe rt.. Time to double: Solve =e 0.09t, leading to t= ln years. After 5 years: 0.09 S=,500e (0.09)(5) $48,7.8.. APR: Solve =e 4r, leading to r= ln 7.%. 4 After 5 years: S=9500e (0.7)(5) $7,86.6 (using the exact value of r). Section.6 Exercises. A=500(+0.07) 6 $5.0. A=,000(+0.075) 7 $9, A=500a $.7 4 b 7. A=40,500a In #5 40, the time must be rounded up to the end of the 40 $86,496.6 b next compounding period (except in the case of continuous compounding). 9. A=50e (0.054)(6) $ Solve a t round. A=,000e (0.07)(0) $0, b = : t = ln 4 ln.0 L 7.4 a to 7 years 6 months. 4. FV= 500 # b - ln $4, Solve.07 t =: t= round to years ln.07 L a Solve a t 0 b = : t = ln ln( /) b round to 0 years. 5. FV= 450 # $70, For #4 44, observe that the initial balance has no effect on the APY. - a PV= 85.7 # b APY= a b - L 6.4% $4,5. 4. APY=e % (8,000)a PV # i 9. R= $9.4 - ( + i) -n = b 45. The APYs are a and b - L 5.6% - a b a b - L 5.984%. So, the better investment 4 is 5.% compounded quarterly. In # 4, the time must be rounded up to the end of the next compounding period. 8. Solve 00 a For #47 50, use the formula S= R ( + i)n -. 4t (.05) 4t =,so 4 b i = 450: i= = and R=50, so ln(8/46) t= 6.6 years round to 6 years 4 ln.05 L (.00605) ()(5) - S=50 L $4, months (the next full compounding period) Solve 5,000 a t (.0067) t =, so b 0.4 = 45,000: 49. i= solve ; ln t=.7 years round to years ln.0067 L a ( * 0) b - 9 months (the next full compounding period). Note: A 50,000=R to obtain 0.4 graphical solution provides t L.78 years round to years 0 months. R $9.4 per month (round up, since $9.4 will not 5. Solve,000a + r (65)(5) 65 b be adequate). = 6,500: - ( + i) -n + r, so r 0.%. 65 = a 7 /85 For #5 54, use the formula A=R. 44 b i Solve 4.6 (+r) 6 =: +r= a 0 /6 5. i= = ; solve,so 7 b r 7.07%. 9000= R - (.00665)-()(4) to obtain R $ per month. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley. 4t b

14 88 Chapter Exponential, Logistic, and Logarithmic Functions i= = ; solve 86,000= R - (.00797)-()(0) to obtain R $ per month (round up, since $ will not be adequate) (a) With i=, solve = (.0) -n 86,000=050 ; this leads to 0.0 (.0) n = - 860, so n = 9 05 months, or about 4. years. The mortgage will be paid off after 7 months (4 years, 4 months). The last payment will be less than $050. A reasonable estimate of the final payment can be found by taking the fractional part of the computed value of n above, 0.8, and multiplying by $050, giving about $ To figure the exact amount of the final payment, - (.0) -7 solve 86,000=050 +R (.0) (the present value of the first 7 payments, plus the present value of a payment of R dollars 7 months from now). This gives a final payment of R $ (b) The total amount of the payments under the original plan is 60 # $884.6 = $8, The total using the higher payments is 7 # $050 = $80,660 (or 7 # $050 + $ = $80,96.57 if we use the correct amount of the final payment) a difference of $7, (or $8,06.0 using the correct final payment). 57. One possible answer: The APY is the percentage increase from the initial balance S(0) to the end-of-year balance S(); specifically, it is S()/S(0)-. Multiplying the initial balance by P results in the end-of-year balance being multiplied by the same amount, so that the ratio remains unchanged. Whether we start with a $ investment, or a $000 investment, APY= a + r k. k b One possible answer: Some of these situations involve counting things (e.g., populations), so that they can only take on whole number values exponential models which predict, e.g., 49.7 fish, have to be interpreted in light of this fact. Technically, bacterial growth, radioactive decay, and compounding interest also are counting problems for example, we cannot have fractional bacteria, or fractional atoms of radioactive material, or fractions of pennies. However, because these are generally very large numbers, it is easier to ignore the fractional parts. (This might also apply when one is talking about, e.g., the population of the whole world.) Another distinction: while we often use an exponential model for all these situations, it generally fits better (over long periods of time) for radioactive decay than for most of the others. Rates of growth in populations (esp. human populations) tend to fluctuate more than exponential models suggest. Of course, an exponential model also fits well in compound interest situations where the interest rate is held constant, but there are many cases where interest rates change over time. 6. False. The limit, with continuous compounding, is A = Pe rt = 00 e 0.05 L $ A = P( + r/k) kt = 50( /4) 4(6) L $4.00. The answer is B. 65. FV=R((+i) n -)/i= 00(( ) 40 -)/ $6,47.. The answer is E. 67. The last payment will be $64.8. ( + i) n (a) Matching up with the formula S=R, i where i=r/k, with r being the rate and k being the number of payments per year, we find r=8%. (b) k= payments per year. (c) Each payment is R=$00. Chapter Review. fa b = - # 4 / = - 4 For # 4, recall that exponential functions have the form f(x)=a # b x.. a =, so f() = # b f(x)= # = 6, b =, b =, x/ 5. f(x)= x + starting from x, horizontally shrink by, reflect across y-axis, and translate up units. [ 4, 6] by [0, 0] 7. f(x)= x - starting from x, horizontally shrink by, reflect across the y-axis, reflect across x-axis, translate down units. [, ] by [9, ] 9. Starting from e x, horizontally shrink by, then translate right units or translate right units, then horizontally shrink by. [, ] by [0, 0] Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

15 Chapter Review f(0) =,, lim f(x) = =.5 lim f(x) = 0 For # 6, recall that logistic functions are expressed in x: -q c f(x)=. y-intercept: (0,.5). + ae - bx 0 Asymptotes: y=0 and y=0. c=0, a=.5, so f()= =0, +.5e -b. It is an exponential decay function. lim f(x) =, lim f(x) = q 0=0+0e b, 0e b =0, e b =, x: -q b ln e=ln.0986, so b L 0 Thus, f(x) =. +.5e -0.55x 0 5. c=0, a=, so f()= =0, + e -b [ 5, 0] by [ 5, 5] 0 0=0+0e b, 0e b =0, e b =, 0 = 5. b ln e=ln.0986, so b 0.7. L 0 Thus, f(x) L. + e -0.7x 7. log =log 5 =5 log =5 [, 4] by [ 0, 0] 9. log 0log 0 = log 0= Domain: ( q, q). x= 5 =4 Range: (, q). a x Always decreasing y b = e- Not symmetric x = y Bounded below by y=, which is also the only e asymptote 5. Translate left 4 units. lim f(x) =, lim f(x) = q x: -q 7. [ 5, 0] by [, 8] Domain: ( q, q) Range: (0, 6) Increasing Symmetric about (.0, ) Bounded above by y=6 and below by y=0, the two asymptotes No extrema lim f(x) = 6, lim f(x) = 0 x: -q For #9, recall that exponential functions are of the form f(x) = a # ( + r) kx. 9. a=4, r=0.05, k=; so f(x)=4 #.05 x, where x=days.. a=8, r=, k=, so f(x)= 8 # x/, where x=days. 7. Translate right unit, reflect across x-axis, and translate up units. 9. [ 6, 7] by [ 6, 5] [0, 0] by [ 5, 5] [ 4.7, 4.7] by [.,.] Domain: (0, q) Range: c - [ 0.7, q) e, q b L Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

16 90 Chapter Exponential, Logistic, and Logarithmic Functions 4. Decreasing on (0, 0.7]; increasing on [0.7, q) Not symmetric Bounded below Local minimum at a e, - e b lim f(x) = q # 7. P(t)= t, where x is the number of years since 900. In 005, P(05)=.0956 # million. years since 900. In 00, P(0).7 million. 7. (a) f(0)=90 units. (b) f().87 units. (c) [ 4.7, 4.7] by [.,.] Domain: ( q, 0) ª (0, q) Range: [ 0.8, q) Discontinuous at x=0 Decreasing on ( q, 0.6], (0, 0.6]; Increasing on [ 0.6, 0), [0.6, q) Symmetric across y-axis Bounded below Local minima at ( 0.6, 0.8) and (0.6, 0.8) No asymptotes lim f(x) = q, lim f(x) = q x: -q 4. x=log ln 45. x=.57 ln x=0 7 = log x=, so x= =4 5. Multiply both sides by # x, leaving ( x ) -=0 # x, or ( x ) -0 # x -=0. This is quadratic in x, leading 0 ; to x = = 5 ; 6. Only 5+ 6 is positive, so the only answer is x=log (5+ 6) log[(x+)(x-)]=4, so (x+)(x-)=0 4. The solutions to this quadratic equation are x= ( ; 40,009), but of these two numbers, only the positive one, x = (40,009 - ) L 99.5, works in the original equation. ln x 55. log x= ln log x 57. log 5 x= log Increasing, intercept at (, 0). The answer is (c). 6. Intercept at (, 0). The answer is (b). 6. A=450(+0.046) $ A=Pe rt 550a - a (-)(5) b b 67. PV = $8, a b 69. 0e k =50, so k= ln 5 L [0, 4] by [0, 90] 75. (a) P(t)=89,000(-0.08) t =89,000(0.98) t. ln(50/89) (b) P(t)=50,000 when t=.74 years. ln (a) P(t)=0 # t, where t is time in months. (Other possible answers: 0 # t if t is in years, or 0 # t/0 if t is in days). (b) P()=8,90 rabbits after year. P(60) rabbits after 5 years. (c) Solve 0 # t =0,000 to find t=log months 8 months and about 9 days. 79. (a) S(t)=S 0 # a t/.5, where t is time in seconds. b (b) S(.5)=S 0 /. S()=S 0 /4. (c) If g=s(60)=s 0 =S0 # a 40 # a 60/.5, then b b S 0 = *0 g=.0995*0 9 kg =,099,500 metric tons. 8. Let a =the amplitude of the ground motion of the Feb 4 quake, and let a =the amplitude of the ground motion of the May 0 quake. Then: 6. = log a T + B and 6.9 = log a T + B alog a T + Bb - alog a + Bb = T a log T - log a T = 0.8 log = 0.8 a = a a L 6. a. The ground amplitude of the deadlier quake was approximately 6. times stronger. 8. Solve 500 a t =750: (.0) 4t =.5, 4 b ln.5 so t=.5678 years round to years 4 ln.0 L 9 months (the next full compounding period) t=.8 ln about years 6 months. 50 L 87. r= a b % 89. I= # 0 ( 0.05)(5) =5.84 lumens a a Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.

17 Chapter Project 9 log x 9. log b x=. This is a vertical stretch if <b<0 log b 0 (so that log b <), and a shrink if 0<b< or 0 b>0. (There is also a reflection if 0<b<.) 9. (a) P(0)=6 students. (b) P(t)=800 when +99e 0.4t =, or e 0.4t =99, so t= ln about days. 0.4 (c) P(t)=400 when +99e 0.4t =4, or e 0.4t =, so t= ln 8.74 about 8 or 9 days The model is T=0+76e kt, and T(8)=65 45 =0+76e 8k. Then e 8k =, so k= - 45 ln Finally, T=5 when 5=0+76e kt, so t= minutes. k ln 5 76 L ( + i) n (a) Matching up with the formula S=R, i where i=r/k, with r being the rate and k being the number of payments per year, we find r=9%. (b) k=4 payments per year. (c) Each payment is R=$ (a) Grace s balance will always remain $000, since interest is not added to it. Every year she receives 5% of that $000 in interest; after t years, she has been paid 5t% of the $000 investment, meaning that altogether she has # 0.05t=000(+0.05t). (b) The table is shown below; the second column gives values of 000e 0.05t. The effects of compounding continuously show up immediately. Not Years Compounded Compounded Chapter Project Answers are based on the sample data shown in the table.. Writing each maximum height as a (rounded) percentage of the previous maximum height produces the following table. Bounce Number Percentage Return 0 N/A 79% 77% 76% 4 78% 5 79% The average is 77.8%. 4. Each successive height will be predicted by multiplying the previous height by the same percentage of rebound. The rebound height can therefore be predicted by the equation y=hp x where x is the bounce number. From the sample data, H=.788 and P y=hp x becomes y.788 # x. 6. The regression equation is y.7 # x. Both H and P are close to, though not identical with, the values in the earlier equation. 7. A different ball could be dropped from the same original height, but subsequent maximum heights would in general change because the rebound percentage changed. So P would change in the equation. 8. H would be changed by varying the height from which the ball was dropped. P would be changed by using a different type of ball or a different bouncing surface. 9. y=hp x =H(e ln P ) x (ln P)x =He =.788 e 0.5x 0. ln y=ln (HP x ) =ln H+x ln P This is a linear equation.. [, 6] by [0, ] Bounce Number ln (Height) [, 6] by [.5,.5] The linear regression produces Y=ln y 0.5x Since ln y (ln P)x+ln H, the slope of the line is ln P and the Y-intercept (that is, the ln y-intercept) is ln H. Copyright 0 Pearson Education, Inc. Publishing as Addison-Wesley.