Chapter 3 Exponential, Logistic, and Logarithmic Functions

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Section 3.1 Exponential and Logistic Functions 133

■ Section 3.1 Exponential and Logistic 2. The point (0, 1) is common to all four graphs, and all four Functions functions can be described as follows: Domain: (- q, q) Range: (0, q) Exploration 1 Continuous 1. The point (0, 1) is common to all four graphs, and all four Always decreasing functions can be described as follows: Not symmetric - q q Domain: ( , ) No local extrema q Range: (0, ) Bounded below by y = 0 , which is also the only Continuous asymptote Always increasing lim g(x) = 0, lim g(x) = q Not symmetric x: q x:–q No local extrema Bounded below by y = 0 , which is also the only asymptote lim f(x) = q, lim f(x) = 0 1 x x: q x:–q y = 1 a 2 b

[–2, 2] by [–1, 6] x y1=2

[–2, 2] by [–1, 6] 1 x y = 2 a 3 b

[–2, 2] by [–1, 6] x y2=3

[–2, 2] by [–1, 6] 1 x y = 3 a 4 b

x y3=4 [–2, 2] by [–1, 6]

[–2, 2] by [–1, 6] 1 x y = 4 a 5 b

x y4=5 [–2, 2] by [–1, 6]

[–2, 2] by [–1, 6]

134 Chapter 3 Exponential, Logistic, and Logarithmic Functions

Exploration 2 1 6. 8 1. 3 1 7. a6 15 f x = 2x 8. b 1 2 9. –1.4, since (–1.4)5=–5.37824 10. 3.1, since (3.1)4=92.3521 [–4, 4] by [–2, 8] 3.1 2. Section Exercises 1. Not an exponential function because the base is variable and the exponent is constant. It is a monomial function. f x = 2x 2. Exponential function, with an initial value of 1 and base of 3. 1 2 g x = e0.4x 3. Exponential function, with an initial value of 1 and base 1 2 of 5. [–4, 4] by [–2, 8] 4. Not an exponential function because the exponent is constant. It is a constant function. 5. Not an exponential function because the base is variable. 6. Not an exponential function because the base is variable. f x = 2x It is a power function. 1 2 # 0 # g x = e0.5x 7. f(0) = 3 5 = 3 1 = 3 1 2 6 2 8. f(-2) = 6 # 3-2 = = [–4, 4] by [–2, 8] 9 3 1 9. f =-2 # 31/3 =-223 3 a 3 b

3 # -3/2 8 8 8 f x = 2x 10. f(- ) = 8 4 = = = = 1 1 2 2 (22)3/2 23 8 g x = e0.6x 1 2 3 1 x 11. f(x) = # 2 a 2 b [–4, 4] by [–2, 8] 1 x 12. g(x) = 12 # a 3 b 13. f(x) = 3 # (22)x = 3 # 2x/2 f x = 2x 1 x 1 2 14. g(x) = 2 # = 2e-x g x = e0.7x a e b 1 2 15. Translate f(x) = 2x by 3 units to the right. Alternatively, [–4, 4] by [–2, 8] - - 1 1 g(x) = 2x 3 = 2 3 # 2x = # 2x = # f(x), so it can be 8 8 1 obtained from f(x) using a vertical shrink by a factor of . 8 f x = 2x 1 2 g x = e0.8x 1 2

[–4, 4] by [–2, 8] k = 0.7 most closely matches the graph of f(x). 3. k L 0.693 [–3, 7] by [–2, 8] 16. Translate f(x)=3x by 4 units to the left. Alternatively, Quick Review 3.1 g(x)=3x±4=34 # 3x=81 # 3x=81 # f(x), so it can be 1. 23 -216 =-6 since (-6)3 =-216 obtained by vertically stretching f(x) by a factor of 81. 125 5 2. = since 53 = 125 and 23 = 8 B3 8 2 3. 272/3=(33)2/3=32=9 4. 45/2=(22)5/2=25=32 1 5. 12 2 [–7, 3] by [–2, 8] Copyright © 2015 Pearson Education, Inc. ch03_p133_163.qxd 1/22/14 1:41 PM Page 135

Section 3.1 Exponential and Logistic Functions 135

17. Reflect f(x) = 4x over the y-axis. 23. Reflect f(x) = ex across the y-axis, horizontally shrink by a factor of 3, translate 1 unit to the right, and vertically stretch by a factor of 2.

[–2, 2] by [–1, 9] 18. Reflect f(x) = 2x over the y-axis and then shift by 5 units [–2, 3] by [–1, 4] to the right. 24. Horizontally shrink f(x) = ex by a factor of 2, vertically stretch by a factor of 3, and shift down 1 unit.

[–3, 7] by [–5, 45] 19. Vertically stretch f(x) = 0.5x by a factor of 3 and then shift 4 units up. [–3, 3] by [–2, 8] 25. Graph (a) is the only graph shaped and positioned like the graph of y = bx, b 7 1. 26. Graph (d) is the reflection of y = 2x across the y-axis. 27. Graph (c) is the reflection of y = 2x across the x-axis. 28. Graph (e) is the reflection of y = 0.5x across the x-axis. 29. Graph (b) is the graph of y = 3-x translated down 2 units. [–5, 5] by [–2, 18] 30. Graph (f) is the graph of y = 1.5x translated down 2 units. 20. Vertically stretch f(x) = 0.6x by a factor of 2 and then 31. Exponential decay; lim f(x) = 0; lim f(x) = q horizontally shrink by a factor of 3. x: q x:–q 32. Exponential decay; lim f(x) = 0; lim f(x) = q x: q x:–q 33. Exponential decay: lim f(x) = 0; lim f(x) = q x: q x:–q 34. Exponential growth: lim f(x) = q; lim f(x) = 0 x: q x:–q 35. x 6 0 [–2, 3] by [–1, 4] 21. Reflect f(x) = ex across the y-axis and horizontally shrink by a factor of 2.

[–2, 2] by [–0.2, 3]

36. x 7 0

[–2, 2] by [–1, 5] 22. Reflect f(x) = ex across the x-axis and y-axis. Then, horizontally shrink by a factor of 3.

[–0.25, 0.25] by [0.5, 1.5]

[–3, 3] by [–5, 5]

136 Chapter 3 Exponential, Logistic, and Logarithmic Functions

37. x 6 0 45.

[–3, 3] by [–2, 8] [–0.25, 0.25] by [0.75, 1.25] Domain: (- q, q) Range: (0, q) 38. x 7 0 Continuous Always increasing Not symmetric Bounded below by y = 0 , which is also the only asymptote No local extrema lim f(x) = q, lim f(x) = 0 x: q x:–q 46. [–0.25, 0.25] by [0.75, 1.25]

= 2x+4 = 2(x+2) = 2 x+2 = x+2 39. y1 y3, since 3 3 (3 ) 9 . = # 3x–2 = 1 3x–2 = 1+3x–2 = 3x–1 40. y2 y3, since 2 2 2 2 2 2 . 41. y-intercept: (0, 4). Horizontal asymptotes: y=0, y = 12. [–3, 3] by [–2, 18] Domain: (- q, q) Range: (0, q) Continuous Always decreasing Not symmetric [Ð10, 20] by [Ð5, 15] Bounded below by y = 0 , which is the only asymptote 42. y-intercept: (0, 3). Horizontal asymptotes:y = 0 ,y = 18 . No local extrema lim f(x) = 0, lim f(x) = q x: q x:–q 47.

[Ð5, 10] by [Ð5, 20]

43. y-intercept: (0, 4). Horizontal asymptotes:y = 0 ,y = 16 . [–2, 2] by [–1, 9] Domain: (- q, q) Range: (0, q) Continuous Always increasing Not symmetric Bounded below by y = 0 , which is the only asymptote No local extrema [Ð5, 10] by [Ð5, 20] lim f(x) = q, lim f(x) = 0 : q : q 44. y-intercept: (0, 3). Horizontal asymptotes:y = 0 ,y = 9 . x x –

[Ð5, 10] by [Ð5, 10]

Section 3.1 Exponential and Logistic Functions 137

48. 52. Let P(t) be Columbus’s population t years after 1990. = # t Then with exponential growth,P(t) P0 b where P0=632,910. From Table 3.7, P(21)=632,910 . b21=7797,434. So, 797,434 b = L 1.0111. B21 632,910 [–2, 2] by [–1, 9] Solving graphically, we find that the curve y = 632,910 (1.0111)t intersects the line y=800,000 at t L 21.22. Domain: (- q, q) Columbus’s population will pass 800,000 in 2011. Range: (0, q) Continuous 53. Using the results from Exercises 51 and 52, we represent t Always decreasing Austin’s population as y=465,622(1.0274) and t Not symmetric Columbus’s population as y=632,910(1.0111) . Solving L Bounded below by y = 0 , which is also the only asymptote graphically, we find that the curves intersect at t 19.19 . No local extrema The two populations were equal, at 782,273, in 2009. lim f(x) = 0, lim f(x) = q 54. Let P(t) be Austin’s population t years after 1990. Then with x: q x:–q exponential growth,P(t) = P bt where P = 465,622 . 49. 0 0 From Table 3.7,P(21) = 465,622 b21 = 820,611 . So, 820,611 b = L 1.0274. B21 465,622 Solving graphically, we find that the curve y = 465,622(1.0274)t intersects the line y=1,000,000 at t L 28.28. Austin’s population passed 1,000,000 in 2018. [–3, 4] by [–1, 7] Let P(t) be Columbus’s population t years after 1990. Domain: (- q, q) = t Then with exponential growth,P(t) P0 b where Range: (0, 5) P0=632,910. From Table 3.7, Continuous P(21)=632,910 b21=7797,434. So, Always increasing 797,434 Symmetric about (0.69, 2.5) b = L 1.0111. B21 Bounded below by y = 0 and above by y = 5 ; both are 632,910 = asymptotes Solving graphically, we find that the curve y 632,910 t L No local extrema (1.0111) intersects the line y=1,000,000 at t 41.44. lim f(x) = 5, lim f(x) = 0 Columbus’s population will pass 1,000,000 in 2031. : q : q x x – From the results of above, Austin’s population will reach 50. 1,000,00 first, in 2018. 55. Solving graphically, we find that the curve 12.79 y = intersects the line y=10 when 1 + 2.402e-0.0309x t L 69.67. Ohio’s population stood at 10 million in 1969. 19.875 [–3, 7] by [–2, 8] 56. (a) P(50) = L 1.794558 1 + 57.993e-0.035005(50) - q q Domain: ( , ) or 1,794,558 people. Range: (0, 6) 19.875 Continuous (b) P(215) = L 19.272737 or -0.035005(215) Always increasing 1 + 57.993e Symmetric about (0.69, 3) 19,272,737 people. Bounded below by y = 0 and above by y = 6 ; both (c) lim P(t) = 19.875 or 19,875,000 people. x: q are asymptotes 57. (a) When t = 0, B = 100. No local extrema = L lim f(x) = 6, lim f(x) = 0 (b) When t 6, B 6394. : q : q x x – 58. (a) When t = 0, C = 20 grams. For #51 and 52, refer to Example 7 on page 260 in the text. (b) When t = 10,400, C L 5.647 . After about 5700.22 years, 51. Let P(t) be Austin’s population t years after 1990. Then with 10 grams remain. # t exponential growth,P(t) = P b where P = 465,622 . 0 0 59. False. If a>0 and 01, then From Table 3.7, P(21) = 465,622 .b21 = 820,611.So, f(x) = a # bx is decreasing. 820,611 b = L 1.0274. c B21 60. True. For f(x) = the horizontal asymptotes 465,622 1 + a # bx Solving graphically, we find that the curve are y=0 and y=c, where c is the limit of growth. t y = 465,622(1.0274) intersects the line y=800,000 at 61. Only 8x has the form a # bx with a nonzero and b positive t L 20.02. Austin’s population will pass 800,000 in 2010. but not equal to 1. The answer is E.

138 Chapter 3 Exponential, Logistic, and Logarithmic Functions

0 62. For b>0, f(0)=b =1. The answer is C. 68. (a) y1—f(x) decreases less rapidly as x increases. # x 63. The growth factor of f(x) = a b is the base b. The (b) y3—as x increases, g(x) decreases ever more rapidly. answer is A. 69. c = 2a: To the graph of (2a)x apply a vertical stretch by 2b , x x 64. With x>0, a >b requires a>b (regardless of since f(ax + b) = 2ax+b = 2ax2b = (2b)(2a)x. whether x<1 or x>1). The answer is B. 70. a Z 0, c = 2. 65. (a) 71. a 6 0, c = 1. 72. a 7 0 and b 7 1, or a 6 0 and 0 6 b 6 1. 73. a 7 0 and 0 6 b 6 1, or a 6 0 and b 7 1. 74. Since 0

Section 3.2 Exponential and Logistic Modeling 139

12. f(x) = 502,000 # (1 + 0.017)x = 502,000 # 1.017x 60 28. c=60, a=3, so f(8)= = 30, 60 = 30 + 90b8, (x=years) + 8 # # 1 3b 13. f(x) = 18 (1 + 0.052)x = 18 1.052x (x = weeks) 1 1 90b°=30, b°= , b= 8 L 0.87 , 14. f(x) = 15 # (1 - 0.046)x = 15 # 0.954x (x = days) 3 B3 # x/3 60 15. f(x)=0.6 2 (x=days) thus f(x)= . 1 + 3 # 0.87x 16. f(x)=250 # 2x/7.5 = 250 # 22x/15 (x = hours) 29. P(t) = 736,000(1.0149)t; P(t)=1,000,000 when t≠20.73 # -x/6 17. f(x)=592 2 (x=years) years, or in the year 2020. 18. f(x)=17 # 2-x/32 (x=hours) 30. P(t) = 478,000(1.0628)t; P(t)=1,000,000 when t≠12.12 2.875 19. f = 2.3, = 1.25 = r + 1, so years, or in the year 2012. 0 2.3 = t f(x)=2.3 # 1.25x (Growth Model). 31. The model is P(t) 6250(1.0275) . -4.64 (a) In 1915: about P(25)≠12,315. In 1940: about 20. g =-5.8, = 0.8 = r + 1, so P(50)≠24,265. 0 -5.8 g(x)=-5.8 # 0.8x (Decay Model) . (b) P(t)=50,000 when t≠76.65 years after 1890 — in 1966. For #21 and 22, use f(x) = f # bx . 0 = t = = # x = # 5 = 32. The model is P(t) 4200(1.0225) . 21. f0 4, so f(x) 4 b . Since f(5) 4 b 8.05, (a) In 1930: about P(20)≠6554. In 1945: about P(35) 8.05 8.05 bfi=, b=L 1.15. f(x) L 4 # 1.15x . ≠9151. 4 B5 4 (b) P(t)=20,000 when t≠70.14 years after 1910: about = = # x = # 4 = 22. f0 3, so f(x) 3 b . Since f(4) 3 b 1.49 1980. t/14 1.49 1.49 # x 1 b›= , b=4 L 0.84. f(x) L 3 0.84 . 33. (a) y = 6.6 , where t is time in days. 3 B 3 a 2 b c For #23–28, use the model f(x) = . (b) After 38.11 days. 1 + a # bx 40 1 t/65 23. c=40, a=3, so f(1)= = 20, 20 + 60b = 40, 34. (a) y = 3.5 , where t is time in days. 1 + 3b a 2 b 1 40 60b=20, b= , thus f(x)= . (b) After 117.48 days. 3 1 x 1 + 3 # 35. One possible answer: Exponential and linear functions a 3 b are similar in that they are always increasing or always 60 decreasing. However, the two functions vary in how 24. c=60, a=4, so f(1)= = 24, 60 = 24 + 96b, 1 + 4b quickly they increase or decrease. While a linear function 3 60 will increase or decrease at a steady rate over a given 96b=36, b= , thus f(x)= . 8 3 x interval, the rate at which exponential functions increase 1 + 4 a 8 b or decrease over a given interval will vary. 128 36. One possible answer: Exponential functions and logistic 25. c=128, a=7, so f(5)= = 32, 1 + 7b5 functions are similar in the sense that they are always 96 increasing or always decreasing. They differ, however, in 128=32+224bfi, 224bfi=96, bfi= , the sense that logistic functions have both an upper and a 224 lower limit to their growth (or decay), while exponential 96 128 b= L 0.844 , thus f(x) L . functions generally have only a lower limit. (Exponential B5 + # x 224 1 7 0.844 functions just keep growing.) 30 = = + 3 37. One possible answer: From the graph we see that the 26. c=30, a=5, so f(3)= 3 15, 30 15 75b , 1 + 5b doubling time for this model is 4 years. This is the time 15 1 1 required to double from 50,000 to 100,000, from 100,000 to 75b‹=15, b‹== , b= L 0.585 , 75 5 B3 5 200,000, or from any population size to twice that size. 30 Regardless of the population size, it takes 4 years for it to thus f(x) L . 1 + 5 # 0.585x double. 20 38. One possible answer: The number of atoms of a radioac- 27. c=20, a=3, so f(2)= = 10, 20 = 10 + 30b2, 1 + 3b2 tive substance that change to a nonradioactive state in a given time is a fixed percentage of the number of radioac- 1 1 30b¤=10, b¤= , b= L 0.58, tive atoms initially present. So the time it takes for half of B 3 3 the atoms to change state (the half-life) does not depend 20 thus f(x)= . on the initial amount. 1 + 3 # 0.58x 39. When t=1, B≠200 — the population doubles every hour. 40. The half-life is about 5700 years.

140 Chapter 3 Exponential, Logistic, and Logarithmic Functions

For #41 and 42, use the formula P(h)=14.7 # 0.5h/3.6, where h 1,301,642 48. P(t) L , which is the same model as -0.05054t is miles above sea level. 1 + 21.602 e 41. P(10)=14.7 # 0.510/3.6=2.14 lb/in2 the solution in Example 8 of Section 3.1. Note that t represents the number of years since 1900. 42. P(h) = 14.7 # 0.5h/3.6 intersects y = 2.5 when h≠9.20 miles above sea level.

[0,120] by [0,1500000] 19.875 [–1, 19] by [–1, 9] 49. P(x) L where x is the number of 1 + 57.993e-0.035005x 43. The exponential regression model is P(t) = 1149.61904 years after 1800 and P is measured in millions. Our model (1.012133)t, where P(t) is measured in thousands of people is the same as the model in Exercise 56 of Section 3.1. and t is years since 1900. The predicted population for Los Angeles for 2011 is P(111) L 4384.5, or 4,384,500 people. This is an overestimate of 565,000 people, 565,000 an error of L 0.15 = 15% . 3,820,000 44. The exponential regression model using 1950–2000 data is P(t) = 20.84002(1.04465)t, where P(t) is measured in [0, 200] by [0, 20] thousands of people and t is years since 1900. The predicted 15.64 50. P(x) L , where x is the number of population for Phoenix for 2011 is P(111) L 2658.6, or 1 + 11799.36e-0.043241x 2,658,600 people. This is an overestimate of 1,189,200 people, years since 1800 and P is measured in millions. 1,189,200 As x : q, P(x) : 15.64, or nearly 16 million, which is an error of L 0.81 = 81% . 1,469,4000 significantly less than New York’s population limit of 20 million. The population of Arizona, according to our The equations in #45 and 46 can be solved either algebraically models, will not surpass the population of New York. Our or graphically; the latter approach is generally faster. graph confirms this. 45. (a) P(0)=16 students. (b) P(t)=200 when t≠13.97 — about 14 days. (c) P(t)=300 when t≠16.90 — about 17 days. 46. (a) P(0)=11. (b) P(t)=600 when t≠24.51 — after 24 or 25 years. (c) As t : q ,P(t) : 1001 —the population never rises above this level. [0, 500] by [0, 25] 47. The logistic regression model is 51. False. This is true for logistic growth, not for exponential growth. 758.4018732 P(x) = , where x is the 52. False. When r<0, the base of the function, 1+r, is + -0.01626448x 1 8.69834271e merely less than 1. number of years since 1900 and P(x) is measured in mil- 53. The base is 1.049=1+0.049, so the constant percentage = lions of people. In the year 2020,x 120 , so the model growth rate is 0.049=4.9%. The answer is C. predicts a population of P(120) L 339.3 or about 54. The base is , so the constant percentage 339.3 million people. 0.834=1-0.166 decay rate is 0.166=16.6%. The answer is B. 55. The growth can be modeled as P(t)=1 # 2t/4. Solve P(t)=1000 to find t≠39.86. The answer is D. 56. Check S(0), S(2), S(4), S(6), and S(8). The answer is E. 694.27 57. (a) P(x) L , where x is the number of 1 + 7.90e-0.017x years since 1900 and P is measured in millions. [Ð10, 130] by [Ð20, 400] P(111)≠308.9, or about 308,900,000 people. (b) The logistic model underestimates the 2000 population by about 2.7 million, an error of around 0.8%. (c) The logistic model predicted a value closer to the actual value than the exponential model, perhaps indicating a better fit.

Section 3.3 Logarithmic Functions and Their Graphs 141

58. (a) Using the exponential growth model and the data e-x - e-(-x) e-x - ex (b) tanh(–x)= = from 1900–2011, Mexico’s population can be repre- -x + -(-x) -x + x x e e e e sented by M(x) L 11.59 * 1.021 , where x is the - ex - e x number of years since 1900 and M is measured in mil- - -tanh( ), so the function is odd. ==x + -x x ions. Using the 1900–2011 data for the United States, e e - and the exponential growth model, the population of ex - e x (c) f(x)=1+tanh(x)=1+ the United States can be represented by P(x) L 81.27 ex + e-x - - * 1.013x,where x is the number of years since 1900 ex + e x + ex - e x 2ex = = and P is measured in millions. Since Mexico’s rate of ex + e-x ex + e-x growth outpaces the United States’ rate of growth, ex 2 2 = - - = , the model predicts that Mexico will eventually have a ex a 1 + e x e x b 1 + e-2x larger population. Our graph indicates that this will which is a logistic function of c=2, a=1, and k=2. occur at x L 233 , or 2133.

■ Section 3.3 Logarithmic Functions and Their Graphs

Exploration 1 1. [Ð20, 280] by [Ð200, 2000] (b) Using logistic growth models and the same data, 159.68 M(x) = 1 + 43.13e-0.0429x 758.40 P(x) = . 1 + 8.698e-0.016x Using this model, Mexico’s population will not exceed that of the United States, confirmed by our graph.

[–6, 6] by [–4, 4] 2. Same graph as part 1.

[Ð20, 500] by [Ð100, 1000] Quick Review 3.3 (c) According to the logistic growth models, the maximum 1 1 1.= 0.04 2. = 0.001 sustainable populations are: 25 1000 Mexico—160 million people. 1 1 United States—758 million people. 3.= 0.2 4. = 0.5 5 2 (d) Answers will vary. However, a logistic model acknowl- 233 326 edges that there is a limit to how much a country’s 5.= 25 = 32 6. = 32 = 9 population can grow. 228 324 1/2 1/3 - - - - 7. 5 8. 10 e x - e ( x) e x - ex 59. sinh(-x) = = 1/2 1/3 2 2 1 = -1/2 1 = -2/3 9.e 10. 2 e ex - e-x a e b a e b =- =-sinh(x), so the function is odd. a 2 b -x + -(-x) -x + x x + -x 3.3 - = e e = e e = e e Section Exercises 60. cosh( x) 1 2 2 2 1. log4 4=1 because 4 =4 =cosh(x), so the function is even. 0 2. log6 1=0 because 6 =1 x -x e - e 5 3. log2 32=5 because 2 =32 sinh(x) 2 4 61. (a) = 4. log3 81=4 because 3 =81 cosh (x) ex + e-x 2 3 = 2/3 3 2 5. log5 225 because 5 =225 - - 3 ex - e x # 2 ex - e x == - - = tanh(x). 2 ex + e x ex + e x 1 =-2 -2/5 = 1 = 1 6. log6 because 6 25 36 5 62/5 25 36

142 Chapter 3 Exponential, Logistic, and Logarithmic Functions

7. log 103=3 42. Starting from y=ln x: translate up 2 units. 8. log 10,000=log 104=4 9. log 100,000=log 105=5 10. log 10–4=–4 1 11. log 23 10=log 101/3= 3

1 –3/2 3 [Ð5, 5] by [Ð3, 4] 12. log =log 10 =- 21000 2 43. Starting from y=ln x: reflect across the y-axis and 13. ln e3=3 translate up 3 units. 14. ln e–4=–4 1 15. ln =ln e–1=–1 e 16. ln 1=ln e0=0 1 17. ln 24 e=ln e1/4= 4 [–4, 1] by [–3, 5] 1 -7/2 7 18. ln = ln e =- 2e7 2 44. Starting from y=ln x: reflect across the y-axis and translate down 2 units. 19. 3, because blogb3 = 3 for any b>0 20. 8, because blogb8 = 8 for any b>0 21. 10log (0.5) = 10log10 (0.5) = 0.5 22. 10log 14 = 10log10 14 = 14 23. eln 6==eloge 6 6 24. eln (1/5)=eloge(1/5) = 1/5 25. log 9.43≠0.9745≠0.975 and 100.9745≠9.43 [–4, 1] by [–5, 1] 26. log 0.908≠–0.042 and 10–0.042≠0.908 45. Starting from y=ln x: reflect across the y-axis and 27. log (–14) is undefined because –14<0. translate right 2 units. 28. log (–5.14) is undefined because –5.14<0. 29. ln 4.05≠1.399 and e1.399≠4.05 30. ln 0.733≠–0.311 and e–0.311≠0.733 31. ln (–0.49) is undefined because –0.49<0. 32. ln (–3.3) is undefined because –3.3<0. 33. x=102=100 34. x=104=10,000 [–7, 3] by [–3, 3] 1 46. Starting from y=ln x: reflect across the y-axis and 35. x=10–1==0.1 10 translate right 5 units. 1 36. x=10–3==0.001 1000 37. f(x) is undefined for x>1. The answer is (d). 38. f(x) is undefined for x<–1. The answer is (b). 39. f(x) is undefined for x<3. The answer is (a). 40. f(x) is undefined for x>4. The answer is (c). [–6, 6] by [–4, 4] 41. Starting from y=ln x: translate left 3 units. 47. Starting from y=log x: translate down 1 unit.

[Ð5, 5] by [Ð3, 3] [Ð5, 15] by [Ð3, 3]

Section 3.3 Logarithmic Functions and Their Graphs 143

48. Starting from y=log x: translate right 3 units. Domain: (2, q) Range: (–q, q) Continuous Always increasing Not symmetric Not bounded No local extrema Asymptote at x=2 [Ð5, 15] by [Ð3, 3] lim f(x)=q x: q 49. Starting from y=log x: reflect across both axes and 54. vertically stretch by 2.

[–2, 8] by [–3, 3] Domain: (–1, q) [–8, 1] by [–2, 3] Range: (–q, q) Continuous 50. Starting from y=log x: reflect across both axes and Always increasing vertically stretch by 3. Not symmetric Not bounded No local extrema Asymptote: x=–1 lim f(x)=q x: q 55.

[–8, 7] by [–3, 3]

51. Starting from y=log x: reflect across the y-axis, translate right 3 units, vertically stretch by 2, translate down 1 unit. [–2, 8] by [–3, 3] Domain: (1, q) Range: (–q, q) Continuous Always decreasing Not symmetric Not bounded [–5, 5] by [–4, 2] No local extrema 52. Starting from y=log x: reflect across both axes, translate Asymptote: x=1 lim f(x) =-q right 1 unit, vertically stretch by 3, translate up 1 unit. x: q 56.

[–6, 2] by [–2, 3] [–3, 7] by [–2, 2] 53. Domain: (–2, q) Range: (–q, q) Continuous Always decreasing Not symmetric Not bounded No local extrema [–1, 9] by [–3, 3] Asymptote: x=–2 lim f(x)=–q x: q Copyright © 2015 Pearson Education, Inc. ch03_p133_163.qxd 1/22/14 1:41 PM Page 144

144 Chapter 3 Exponential, Logistic, and Logarithmic Functions # 57. 62. (a) The exponential regression model is 2552165025 0.995838x, where x is the year and y is the population. (b)

[Ð3, 7] by [Ð3, 3]

Domain: (0, q) [1900, 2100] by [100000, 1000000] Range: (–q, q) Continuous (c) Solving graphically, we find that the curve # x Increasing on its domain y=2552165025 0.995838 intersects the line No symmetry y=500,000 at t=2047. Not bounded (d) Not in most cases as populations will not continue to No local extrema grow without bound. Asymptote at x=0 63. True, by the definition of a logarithmic function. lim f(x) = q 64. True, by the definition of common logarithm. x: q 58. 65. log 2≠0.30103. The answer is C. 66. log 5≠0.699 but 2.5 log 2≠0.753. The answer is A. 67. The graph of f(x)=ln x lies entirely to the right of the origin. The answer is B. = # x -1 = 68. For f(x) 2 3 , f (x) log3(x/2) -1 = # x because f (f(x)) log3(2 3 /2) x [Ð7, 3, 1] by [Ð10, 10, 2] =log3 3 Domain: (–q,2) =x. Range: (–q, q) The answer is A. Continuous 69. f(x) 3x log x Decreasing on its domain ` ` 3 No symmetry Domain (–q, q) (0, q) Not bounded ` ` No local extrema Range (0, q) (–q, q) Asymptote at x=2 ` ` = q limq f(x) Intercepts (0, 1) (1, 0) x: - ` ` 10-11 59. (a) ı=10 log = 10 log 10 = 10(1) =10 dB Asymptotes y=0 x=0 a 10-12 b ` ` 10-5 (b) ı=10 log = 10 log 107 = 10(7) =70 dB a 10-12 b 103 (c) ı=10 log = 10 log 1015 = 10(15) =150 dB a 10-12 b 60. I=12 # 10–0.0705≠10.2019 lumens. 1000 61. (a) A magnitude 3 earthquake is = 10 times more 100 [–6, 6] by [–4, 4] 70. powerful than a magnitude 2 earthquake. A magni- x 100,000 f(x) 5 log5 x tude 5 earthquake is = 100 times more pow- ` ` 1000 Domain (–q, q) (0, q) erful than a magnitude 2 earthquake. ` ` (b) Range (0, q) (–q, q) ` ` Intercepts (0, 1) (1, 0) ` ` Asymptotes y=0 x=0 ` ` [0, 6] by [0, 110000] (c) Ground motion = 10x where x is the magnitude of the earthquake, so y=10x. (d) y=10x, so x=log y. (e) Extremely large values can be represented by much smaller values. (f) Yes [–6, 6] by [–4, 4] Copyright © 2015 Pearson Education, Inc. ch03_p133_163.qxd 1/22/14 1:41 PM Page 145

Section 3.4 Properties of Logarithmic Functions 145

71. b=2 e e. The point that is common to both graphs is (e, e). 4. log 10–3=–3 x5y-2 5. = x5–2y–2–(–4)=x3y2 x2y-4 u-3v7 v7-2 v5 6. - = = u 2v2 u-2-(-3) u |x|3 7. (x6y–2)1/2=(x6)1/2(y–2)1/2= [–3.7, 5.7] by [–2.1, 4.1] |y| 9 72. 0 is not in the domain of the logarithm functions because –8 12 3/4 –8 3/4 12 3/4 |y| x 8. (x y ) =(x ) (y ) = 0 is not in the range of exponential functions; that is, a is x6 never equal to 0. - - (u2v 4)1/2 |u||v| 2 1 73. Reflect across the x-axis. 9. = - = (27 u6v-6)1/3 3u2v 2 3|u| 74. Reflect across the x-axis. (x-2y3)-2 x4y-6 13 = = x 10. 3 -2 -3 - 12 ■ Section 3.4 Properties of Logarithmic (x y ) x 9y6 y Functions Section 3.4 Exercises 1. ln 8x=ln 8+ln x=3 ln 2+ln x Exploration 1 2. ln 9y=ln 9+ln y=2 ln 3+ln y 1. log ()≠2 # 4 0.90309, 3 log 2+log 4≠0.30103+0.60206≠0.90309 3. log =log 3-log x 8 x 2. log ≠0.60206, log 8-log 2≠0.90309-0.30103 a b 2 2 4. log =log 2-log y ≠0.60206 y 3 5 3. log 2 ≠0.90309, 3 log 2≠3(0.30103)≠0.90309 5. log2 y =5 log2 y –2 10 6. log2 x =–2 log2 x 4. log 5=log =log 10-log 2≠1-0.30103 a 2 b 7. log x3y2=log x‹+log y¤=3 log x+2 log y =0.69897 8. log xy3=log x+log y‹=log x+3 log y 4 5. log 16=log 2 =4 log 2≠1.20412 x2 log 32=log 25=5 log 2≠1.50515 9. ln = ln x¤-ln y‹=2lnx-3 ln y y3 log 64=log 26=6 log 2≠1.80618 10. log 1000x4=log 1000+log x4=3+4 log x 2 10 6. log 25=log 5 =2 log 5=2 log x 1 1 1 a 2 b 11. log 4 = (log x-log y)= log x - log y =2(log 10-log 2)≠1.39794 By 4 4 4 log 40 = log (4 # 10) = log 4 + log 10 L 1.60206 23 x 1 1 1 12. ln = (ln x-ln y)= ln x - ln y 100 23 y 3 3 3 log 50 = log = log 100 - log 2 L 1.69897 a 2 b 13. log x+log y=log xy 14. log x+log 5=log 5x The list consists of 1, 2, 4, 5, 8, 16, 20, 25, 32, 40, 50, 64, and 80. 15. ln y-ln 3=ln(y/3) 16. ln x-ln y=ln(x/y) Exploration 2 1 1. False 17. log x=log x1/3=log 23 x 3 2. False; log£ (7x)=log 7+log x 3 3 1 3. True 18. log z=log z1/5=log 25 z 5 4. True 19. 2 ln x+3 ln y=ln x2+ln y3=ln (x2y3) x 5. False; log =log x-log 4 y4 4 20. 4 log y-log z=log y4-log z=log 6. True a z b 2 21. 4 log 3 log log 4 4 log 3 3 7. False; log5 x =log5 x+log5 x=2 log5 x (xy)- (yz)= (x y )- (y z ) 4 4 4 8. True x y x y =log = log a y3z3 b a z3 b Quick Review 3.4 22. 3 ln (x3y)+2 ln (yz2)=ln (x9y3)+ln (y2z4) 1. log 102 2 = =ln (x9y5z4) 2. ln e3=3 3. ln e–2=–2

146 Chapter 3 Exponential, Logistic, and Logarithmic Functions

In #23–28, natural logarithms are shown, but common (base-10) 40. Starting from g(x)=ln x: vertically shrink by a factor logarithms would produce the same results. 1/ln 7≠0.51. ln 7 23. L 2.8074 ln 2 ln 19 24. L 1.8295 ln 5 ln 175 25. L 2.4837 ln 8 [Ð1, 10] by [Ð2, 2] ln 259 26. L 2.2362 ln 12 41. Starting from g(x)=ln x: reflect across the x-axis, then vertically shrink by a factor 1/ln 3≠0.91. ln 12 ln 12 27. =- L-3.5850 ln 0.5 ln 2 ln 29 ln 29 28. =- L-2.0922 ln 0.2 ln 5 ln x 29. log x= 3 ln 3 ln x [Ð1, 10] by [Ð2, 2] 30. log x= 7 ln 7 42. Starting from g(x)=ln x: reflect across the x-axis, then ln(a + b) shrink vertically by a factor of 1/ln 5 L 0.62 . 31. log (a+b)= 2 ln 2

ln(c - d) 32. log (c-d)= 5 ln 5

log x 33. log2 x= log 2 [Ð1, 10] by [Ð2, 2] log x 43. (b): [–5, 5] by [–3, 3], with Xscl=1 and Yscl=1 34. log x= 4 log 4 (graph y=ln(2-x)/ln 4). 44. (c): [–2, 8] by [–3, 3], with Xscl=1 and Yscl=1 log(x + y) log(x + y) =- (graph y=ln(x-3)/ln 6). 35. log1/2(x+y)= log (1/2) log 2 45. (d): [–2, 8] by [–3, 3], with Xscl=1 and Yscl=1 log(x - y) log(x - y) (graph y=ln(x-2)/ln 0.5). 36. log (x-y)= =- 1/3 log(1/3) log 3 46. (a): [–8, 4] by [–8, 8], with Xscl=1 and Yscl=1 (graph y=ln(3-x)/ln 0.7). 37. Let x=log R and y=log S. b b 47. Then bx=R and by=S, so that R bx = = bx-y S by R log = log bx-y = x - y = log R - log S. b a S b b b b

x 38. Let x=logb R. Then# b =R, so that [–1, 9] by [–1, 7] Rc=(bx)c=bc x # Domain: (0, q) log Rc = log bc x = c # x = c log R. b b b Range: (–q, q) 39. Starting from g(x)=ln x: vertically shrink by a factor Continuous 1/ln 4≠0.72. Always increasing Asymptote: x=0 lim f(x)=q x: q ln (8x) f(x)=log (8x)= 2 ln (2)

[–1, 10] by [–2, 2]

Section 3.4 Properties of Logarithmic Functions 147

48. I 53. log = –0.00235(40)=–0.094, so 12 I=12 # 10-0.094 L 9.6645 lumens. I 54. log =–0.0125(10)=–0.125,so 12 I=12 # 10-0.125 L 8.9987 lumens. [–1, 9] by [–5, 2] 55. From the change-of-base formula, we know that Domain: (0, q) = = ln x = 1 # L f(x) log3x ln x 0.9102 ln x. Range: (–q, q) ln 3 ln 3 Continuous f(x) can be obtained from g(x) = ln x by vertically Always decreasing stretching by a factor of approximately 0.9102. Asymptote: x=0 56. From the change-of-base formula, we know that q limq f(x)=– log x 1 x: f(x)=log x= = # log x L-10.32 log x. ln (9x) 0.8 log 0.8 log 0.8 f(x)=log (9x)= 1/3 1 f(x) can be obtained from g(x)=log x by reflecting ln a 3 b across the x-axis and vertically stretching by a factor of approximately 10.32. 49. 57. True. This is the product rule for logarithms. 58. False. The logarithm of a positive number less than 1 is negative. For example, log 0.01=–2. 59. log 12=log (3 # 4)=log 3+log 4 by the product rule. The answer is B.

[–10, 10] by [–2, 3] 60. log9 64=(ln 64)/(ln 9) by the change-of-base formula. The answer is C. q ª q Domain: (– ,0) (0, ) 5 Range: (–q, q) 61. ln x =5 ln x by the power rule. The answer is A. 2 = Discontinuous at x=0 62. log1/2 x 2 log1/2 |x| Decreasing on interval (–q,0); increasing on ln|x| q = 2 interval (0, ) ln(1/2) Asymptote: x=0 q q ln|x| lim f(x)= , lim f(x)= , = 2 x: q x: –q ln 1 - ln 2 ln|x| 50. =-2 ln 2 =- 2 log2 |x|. The answer is E. 63. (a) f(x)=2.75 # x5.0 (b) [–1, 9] by [–2, 8] f(7.1)≠49,616 (c) ln(x) 1.39 1.87 2.14 2.30 Domain: (0, q) p Range: (–q, q) ln(y) 7.94 10.37 11.71 12.53 Continuous Always increasing Asymptote: x=0 lim f(x)=q x: q 51. In each case, take the exponent of 10, add 12, and multiply the result by 10. [0, 3] by [0, 15] (a) 0 (b) 10 (d) ln(y)=5.00 ln x+1.01 (c) 60 (d) 80 (e) a≠5, b≠1 so f(x)=e1x5=ex5≠2.72x5. The two (e) 100 (f) 120 (1 = 100) equations are the same. 250 64. (a) f(x)=8.095 # x–0.113 52. (a) R=log + 4.25 = log 125+4.25≠6.3469 2 (b) f(9.2)=8.095 # (9.2)–0.113≠6.30 300 (b) R=log + 3.5 = log 75+3.5≠5.3751 4

148 Chapter 3 Exponential, Logistic, and Logarithmic Functions

(c) ln(x) 0.69 1.10 1.57 2.04 log27 = log(33) = 3log3 ln(y) p 2.01 1.97 1.92 1.86 log30 = log(10 * 3) = log10 + log3 = 1 + log3 log32 = log(25) = 5log2 log36 = log4 + log9 = 2log2 + 2log3 log40 = log(10 * 4) = log10 + log4 = 1 + 2log2 log45 = log5 + log9 = 1 - log2 + 2log3 log48 = log3 + log16 = log3 + 4log2 100 [0.5, 2.5] by [1.8, 2.1] log50 = log = log100 - log2 = 2 - log2 a 2 b (d) ln(y)=–0.113 ln (x)+2.091 = + = + # log54 log2 log27 log2 3log3 (e) a≠–0.113, b≠2.1, so f(x)=e2.1 x–0.113 log60 = log(10 * 6) = log10 + log6 = 1 + log2 + log3 –0.113 ≠8.09x . log64 = log(26) = 6log2 65. (a) log(w) –0.70 –0.52 0.30 0.70 1.48 1.70 1.85 log72 = log8 + log9 = 3log2 + 2log3 log(r) p 2.62 2.48 2.31 2.08 1.93 1.85 1.86 100 * 3 log75 = log = 2 - 2log2 + log3 a 4 b log80 = log(10 * 8) = log10 + log8 = 1 + 3log2 log81 = log(34) = 4log3 log90 = log(10 * 9) = log10 + log9 = 1 + 2log3 log96 = log3 + log32 = log3 + 5log2 [–1, 2] by [1.6, 2.8] 1 67. Since 25 4581=45811/5 and log (45811/5)= log 4581, (b) log r=(–0.30) log w+2.36 5 we need to find log 4581 L 3.66096 (using tables, before (c) 3.66096 calculators were available) and then find L 5 0.732192, so 25 4581 L 10 0.732192 L 5.40. For #68 and 69, solve graphically. 68. ≠6.41 6 x 6 93.35 69. ≠1.26 … x … 14.77 [–1, 2] by [1.6, 2.8] 70. (a) (d) log r=(–0.30) log (450)+2.36≠1.58, r≠37.69, very close. (e) One possible answer: Consider the power function y=a # xb. Then: log y=log (a # xb) =log a+log xb [–1, 9] by [–2, 8] =log a+b log x q =b(log x)+log a Domain of f and g: (3, ) which is clearly a linear function of the form (b) 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 relation- ship between log y and log x. 66. log4 = 2log2 log5 = log(10/2) = log10 - log2 = 1 - log2 log6 = log2 + log3 [0, 20] by [–2, 8] log8 = log(23) = 3log2 Domain of f and g: (5, q) 2 log9 = log(3 ) = 2log3 (c) log12 = log3 + log4 = log3 + 2log2 10 * 3 log15 = log = log10 - log2 + log3 a 2 b = 1 - log2 + log3 log16 = log(24) = 4log2 log18 = log2 + log9 = log2 + 2log3 [–7, 3] by [–5, 5] log20 = log(10 * 2) = log10 + log2 = 1 + log2 Domain of f: (–q, –3) ª (–3, q) = * = + log24 log(3 8) 3log2 log3 Domain of g: (–3, q) 100 Answers will vary. log25 = log = log100 - log4 = 2 - 2log2 a 4 b Copyright © 2015 Pearson Education, Inc. ch03_p133_163.qxd 1/22/14 1:41 PM Page 149

Section 3.5 Equation Solving and Modeling 149

2 y = (log x )/2 = log x = 71. Recall that y=loga x can be written as x=a . 2. f(g(x)) 10 10 x and x/2 2 = x = Let y=loga b g(f(x))=log(10 ) log(10 ) x. ay=b 1 3x 1 log y log 3. f(g(x)) = ln(e ) = (3x) = x and a = b 3 3 y log a =log b g(f(x))=e3(1/3 ln x) =eln x = x. log b y= = log b. = x/6 2 = x/6 = = log a a 4. f(g(x)) 3 log(10 ) 6 log(10 ) 6(x/6) x 2 and g(f(x)) = 10(3 log x )/6 = 10(6 log x)/6 = 10log x = x. log x 72. Let y= . By the change-of-base formula, * 8 ln x 5. 7.783 10 km log x log e 6. 1 * 10-15 m y== log x # = log e L 0.43 . log x log x 7. 602,000,000,000,000,000,000,000 log e 8. 0.000 000 000 000 000 000 000 000 001 66 (26 zeros Thus, y is a constant function. between the decimal point and the 1). 73. 9. (1.86 * 105)(3.1 * 107) = (1.86)(3.1) * 105+7 =5.766 * 1012 8 * 10-7 8 10. = * 10-7-(-6) = 1.6 * 10-1 5 * 10-6 5 Section 3.5 Exercises [–1, 9] by [–3, 2] For #1–18, take a logarithm of both sides of the equation, Domain: (1, q) when appropriate. Range: (–q, q) 1 x/5 1. 36 = 4 Continuous a 3 b Increasing 1 x/5 1 = Not symmetric a 3 b 9 Vertical asymptote: x=1 1 x/5 1 2 lim f(x)=q = x: q a 3 b a 3 b x One-to one, hence invertible (f–1(x)=ee ) x = 2 5 ■ Section 3.5 Equation Solving and Modeling x = 10 1 x/3 Exploration 1 2. 32 = 2 a 4 b # L 1. log (4 10) 1.60206 1 x/3 1 # 2 L = log (4 10 ) 2.60206 a 4 b 16 # 3 L 1 x/3 1 2 log (4 10 ) 3.60206 = a b a b # 4 L 4 4 log (4 10 ) 4.60206 x = 2 log (4 # 105) L 5.60206 3 x = 6 log (4 # 106) L 6.60206 3. 2 # 5x/4 = 250 # 7 log (4 10 ) L 7.60206 5x/4 = 125 x/4 = 3 log (4 # 108) L 8.60206 5 5 x = # 9 3 log (4 10 ) L 9.60206 4 x = 12 log (4 # 1010) L 10.60206 4. 3 # 4x/2 = 96 2. The integers increase by 1 for every increase in a power 4x/2 = 32 of 10. 4x/2 = 45/2 x 5 3. The decimal parts are exactly equal. = 4. 4 # 1010 is nine orders of magnitude greater than 4 # 10 . 2 2 x = 5 -x/3 = - = =- Quick Review 3.5 5. 10 10, so x/3 1, and therefore x 3. -x/4 = - = =- In #1–4, graphical support (i.e., graphing both functions on a 6. 5 5, so x/4 1, and therefore x 4. square window) is also useful. 7. x = 104 = 10,000 5 1/2 8. x=2 =32 1. f(g(x)) = e2 ln(x ) = eln x = x and g(f(x))=ln(e2x)1/2 - = -1 = + -1 = = ln(ex) = x. 9. x 5 4 , so x 5 4 5.25.

150 Chapter 3 Exponential, Logistic, and Logarithmic Functions

10. 1 - x = 41, so x =-3. 30. Multiply both sides by 2 # 2x, leaving (2x)2 + 1 = 6 # 2x, x 2 - # x + = x ln 4.1 or (2 ) 6 2 1 0. This is quadratic in 2 , leading to 11. x = = log 4.1 L 24.2151 ln 1.06 1.06 6 ; 236 - 4 ln(3 ; 222) 2x = = 3 ; 222. Then x = 2 ln 2 = ln 1.6 = L- 12. x log 0.98 1.6 23.2644 = ; L; ln 0.98 log 2 (3 222) 2.5431. 0.035x 13. e = 4, so 0.035x = ln 4, and therefore 31. Multiply both sides by 2ex, leaving (ex)2 + 1 = 8ex, or 1 ( x)2 - 8 x + 1 = 0. This is quadratic in x, leading to x = ln 4 L 39.6084. e e e 0.035 ; 2 - x 8 64 4 0.045x e = = 4 ; 215 . Then 14. e = 3, so 0.045x = ln 3, and therefore 2 1 x = ln 3 L 24.4136. x=ln(4 ; 215) L;2.0634. 0.045 32. This is quadratic in ex, leading to 3 3 15. e-x = , so -x = ln , and therefore -5 ; 225 + 24 -5 ; 7 2 2 ex = = . Of these two 3 4 4 x =-ln L-0.4055. -5 + 7 1 1 2 numbers, only = is positive, so x = ln 4 2 2 5 5 16. e-x = , so -x = ln , and therefore ≠–0.6931. 3 3 500 3 5 33. = 1 + 25e0.3x, so e0.3x = = 0.06, and therefore x =-ln L-0.5108. 200 50 3 1 1 x = ln 0.06 L-9.3780. 17. ln(x-3)= , so x - 3 = e1/3, and therefore 0.3 3 = + 1/3 L 400 - - 1 x 3 e 4.3956. 34. = 1 + 95e 0.6x, so e 0.6x = , and therefore - 150 57 18. log(x + 2) =-2, so x + 2 = 10 2, and therefore 1 1 x =-2 + 10-2 =-1.99. x = ln L 6.7384. -0.6 57 19. We must have x(x + 1) 7 0, so x 6-1 or x 7 0. - q - q 35. Multiply by 2, then combine the logarithms to obtain Domain:( , 1) h (0, ) ; graph (e). + + x 3 x 3 0 2 20. We must have 7 0 and + 1 7 0, so 7 0. ln = 0. Then = e = 1, so x + 3 = x . x x x x2 x2 Domain:(0, q); graph (f). The solutions to this quadratic equation are x 7 6- 7 1 ; 21 + 12 1 1 21. We must have + 0, so x 1 or x 0. x = = ; 213 L 2.3028. x 1 2 2 2 Domain:(- q, -1) h (0, q) ; graph (d). 36. Multiply by 2, then combine the logarithms to obtain 7 + 7 7 22. We must have x 0 and x 1 0, so x 0. x2 x2 Domain:(0, q); graph (c). log = 2. Then = 102 = 100, so x + 4 x + 4 q 23. We must have x 7 0. Domain:(0, ); graph (a). x2 = 100(x + 4). The solutions to this quadratic equation 2 24. We must have x 7 0, so x Z 0. 100 ; 210000 + 1600 - q q are x = = 50 ; 10229. The Domain:( , 0) h (0, ) ; graph (b). 2 For #25–38, algebraic solutions are shown (and are generally original equation requires that x 7 0, so 50 - 10229 is the only way to get exact answers). In many cases solving extraneous; the only actual solution is graphically would be faster; graphical support is also useful. x = 50 + 10229 L 103.8517. 2 25. Write both sides as powers of 10, leaving 10log x = 106 , or 37. ln[(x-3)(x+4)]=3 ln 2, so (x-3)(x+4)=8, or x2 = 1,000,000. Then x = 1000 or x =-1000. x2 + x - 20 = 0. This factors to (x-4)(x+5)=0, so 2 26. Write both sides as powers of e, leaving eln x = e4, or x=4 (an actual solution) or x=–5 (extraneous, since x2 = e4. Then x = e2 L 7.389 or x =-e2 L-7.389. x-3 and x+4 must be positive). 4 27. Write both sides as powers of 10, leaving 10log x = 102 , or 38. log[(x-2)(x+5)]=2 log 3, so (x-2)(x+5)=9, 2 + - = x4 = 100. Then x2 = 10, and x =;210. or x 3x 19 0. ln x6 12 -3 ; 29 + 76 3 1 28. Write both sides as powers of e, leaving e = e , or Then x = =- ; 285. The actual x6 = e12. Then x2 = e4, and x =;e 2. 2 2 2 3 1 29. Multiply both sides by 3 # 2x, leaving (2x)2 - 1=12 # 2x, solution is x =- + 285 L 3.1098; since x-2 must x 2 - # x - = x 2 2 or (2 ) 12 2 1 0. This is quadratic in 2 , be positive, the other algebraic solution, 12 ; 2144 + 4 3 1 leading to 2x = = 6 ; 237. Only x=- - 285, is extraneous. 2 2 2 6 + 237 is positive, so the only answer is 39. A $100 bill has the value of 1000, or 10‹, dimes so they ln(6 + 237) differ by an order of magnitude of 3. x = = log (6 + 237) L 3.5949. ln 2 2

Section 3.5 Equation Solving and Modeling 151

# 3 + 40. A 2-kg hen weighs 2000, or 2 10 , grams while a 20-g 48. (a) Stomach acid: –log [H ]=2.0 canary weighs 2 # 10 grams. They differ by an order of log [H+]=–2.0 magnitude of 2. [H+]=10–2.0=1*10–2 + 41. 7-5.5=1.5. They differ by an order of magnitude of 1.5. Blood: –log [H ]=7.4 log [H+]=–7.4 42. 4.1-2.3=1.8. They differ by an order of magnitude [H+]=10–7.4≠3.98*10–8 of 1.8. [H+] of stomach acid -2 = 10 L * 5 43. Given (b) + -7.4 2.51 10 I [H ] of blood 10 = 1 = b1 10 log 95 (c) They differ by an order of magnitude of 5.4. I0 I The equations in #49 and 50 can be solved either algebraically = 2 = b2 10 log 65, or graphically; the latter approach is generally faster. I0 we seek the logarithm of the ratio I1/I2. 49. Substituting known information into - I I T(t)=T + (T - T )e kt leaves T(t)=22+70e–kt. 1 - 2 = - m 0 m 10 log 10 log b1 b2 2 I0 I0 Using T(12)=50=22+70e–12k, we have e–12k= , so I1 I2 5 10 log - log = 95 - 65 1 2 a I I b k=- ln L 0.0764. Solving T(t)=30 yields 0 0 12 5 I 10 log 1 = 30 t≠28.41 minutes. I 2 50. Substituting known information into T(t) I1 + - -kt –kt log = 3 =Tm (T0 Tm)e leaves T(t)=65+285e . I2 Using T(20)=120=65+285e–20k, we have The two intensities differ by 3 orders of magnitude. 11 1 11 e–20k= , so k=– ln ≠0.0823. Solving 44. Given 57 20 57 I = 1 = T(t)=90 yields t≠29.59 minutes. b1 10 log 70 I0 51. (a) I = 2 = b2 10 log 10, I0

we seek the logarithm of the ratio I1/I2. I I 1 - 2 = - 10 log 10 log b1 b2 I0 I0 I I [0, 40] by [0, 80] 10 log 1 - log 2 = 70 - 10 a I0 I0 b (b) I 10 log 1 = 60 I2 I log 1 = 6 I2 The two intensities differ by 6 orders of magnitude. 45. Assuming that T and B are the same for the two quakes, [0, 40] by [0, 80] = - + L # x we have 7.9 log a1 log T B and T(x) 79.47 0.93 - + - = 6.6=log a2 log T B, so 7.9 6.6 1.3 (c) T(0)+10=79.47+10≠89.47°C = 1.3 L =log(a1/a2). Then a1/a2 10 , so a1 19.95a2 — the 52. (a) Mexico City amplitude was about 20 times greater. 46. If T and B were the same, we have = - + = - + 7.2 log a1 log T B and 6.6 log a2 log T B, = 0.6 so 7.2-6.6=0.6=log(a1/a2). Then a1/a2 10 , so L a1 3.98a2 — Kobe’s amplitude was about 4 times greater. [0, 35] by [0, 90] 47. (a) Carbonated water: –log [H+]=3.9 log [H+]=–3.9 (b) [H+]=10–3.9≠1.26*10–4 Household ammonia: –log [H+]=11.9 log [H+]=–11.9 [H+]=10–11.9≠1.26*10–12 + -3.9 [H ] of carbonated water 10 8 (b) = - = 10 [H+] of household ammonia 10 11.9 [0, 35] by [0, 90] ( ) L 79.96 # 0.93x (c) They differ by an order of magnitude of 8. T x (c) T(0)+0≠79.96=79.96°C.

152 Chapter 3 Exponential, Logistic, and Logarithmic Functions

53. (a) 61. 23x–1=32 23x–1=25 3x-1=5 x=2 The answer is B. 62. ln x=–1 [0, 20] by [0, 15] eln x=e–1 1 (b) The scatter plot is better because it accurately represents x = the times between the measurements. The equal spacing e on the bar graph suggests that the measurements were The answer is B. taken at equally spaced intervals, which distorts our per- 63. Given a ceptions of how the consumption has changed over time. = 1 + = R1 log B 8.1 54. Answers will vary. T a2 R = log + B = 6.1, 55. Logarithmic seems best — the scatterplot of (x, y) looks 2 T most logarithmic. (The data can be modeled by we seek the ratio of amplitudes (severities) a1/a2. y=3+2 ln x.) a1 a2 log + B - log + B = R - R a T b a T b 1 2 a a log 1 - log 2 = 8.1 - 6.1 T T a log 1 = 2 a2 a1 = 102=100 [0, 5] by [0, 7] a2 The answer is E. 56. Exponential — the scatterplot of (x, y) is exactly exponential. (The data can be modeled by y=2 # 3x.) 64. As the second term on the right side of the formula = + - -kt T(t) Tm (T0 Tm)e indicates, and as the graph confirms, the model is exponential. The answer is A. 943 65. A logistic regression f(x) = most a 1 + 59.126e-0.04786x b closely matches the data, and would provide a natural “cap” to the population growth at approx. 945,000 [0, 5] by [0, 200] people. (Note: x=number of years since 1900.) 57. Exponential — the scatterplot of (x, y) is exactly = 2027 3 66. The logistic regression model f(x) - exponential. (The data can be modeled by y= # 2x.) a 1 + 12.538e 0.0297xb 2 matches the data well and provides a natural cap of 2.0 million people. (Note: x=number of years since 1900.) 67. (a)

[0, 5] by [0, 30]

58. Linear — the scatterplot of (x, y) is exactly linear (y=2x+3.) [–3, 3] by [0, 10] As k increases, the bell curve stretches vertically. Its height increases and the slope of the curve seems to steepen. (b)

[0, 5] by [0, 13] 59. False. The order of magnitude of a positive number is its common logarithm. = + - -kt [–3, 3] by [0, 1] 60. True. In the formula T(t) Tm (T0 Tm)e , the - As c increases, the bell curve compresses horizontally. term (T - T )e kt goes to zero as t gets large, so that 0 m Its slope seems to steepen, increasing more rapidly to T(t) approaches T . m (0, 1) and decreasing more rapidly from (0, 1). Copyright © 2015 Pearson Education, Inc. ch03_p133_163.qxd 1/22/14 1:41 PM Page 153

Section 3.5 Equation Solving and Modeling 153

68. (a and b)

[–302, 502] by [0, 500] [0, 10] by [–5, 3] 71. Since T L 66.156 and T = 4.5, we have (c) 0 m (66.156 - 4.5)e-kt=61.656 * (0.92770)t 61.656e-kt=61.656 * (0.92770)t 61.656 e-kt= * (0.92770)t X = 80 Y = 226.5 X = 226.5 Y = 80 61.656 -kt * t [–302, 502] by [0, 500] [–302, 502] by [0, 500] e =1 (0.92770) ln e-kt=ln (1 # (0.92770)t) - + t = # x =- + kt=ln (1) ln (0.92770) (d) y1 81.274 1.013 and y2 348.540 79.320 ln x -kt=0 + t ln (0.92770) (e) k= - ln (0.92770) ≠0.075. 72. 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 [–302, 502] by [0, 500] linear regression y=ax+b — we make the same substi- # tution (x : ln x) and find y=a ln x+b, a logarithmic 311.6=81.274 1.013x regression. 311.6 73. One possible answer: We “map” our data so that all points = 1.013x 81.274 (x, y) are plotted as (ln x, ln y). If these “new” points are linear — and thus can be represented by some standard 311.6 (f) ln (1.013x)=ln =ln 311.6-ln 81.274 linear regression y=ax+b — we make the same “map- 81.274 ping”(x : ln x, y : ln y) and find ln y=a ln x+b. xln1.013=ln 311.6-ln 81.274 Using algebra and the properties of algorithms, we have: ln y=a ln x+b ln 311.6 - ln 81.274 x = ¢ ≤L 104.05 eln y=ea ln x+b ln1.013 y=ea ln x # eb u ln xa # b 69. Let =10n, u, v 7 0 =e e v =eb # xa u =c xa, where c = eb, exactly the power regression. log =log 10n v The equations and inequalities in #73–76 must be solved log u-log v=n log 10 graphically — they cannot be solved algebraically. For #77 and 78, algebraic solution is possible, although a graphical log u-log v=n(1)=n approach may be easier. For the initial expression to be true, either u and v are both powers of ten, or they are the same constant k multi- 74. x≠1.3066 plied by powers of 10 (i.e., either u=10k and v=10m or u = a # 10k and v = a # 10m, where a, k, and m are constants). As a result, u and v vary by an order of magnitude n. That is, u is n orders of magnitude greater than v. 70. (a) r cannot be negative since it is a distance. [–1, 5] by [–1, 6] 75. x≠0.4073 or x≠0.9333

[–10, 10] by [–10, 30]

(b) [0, 10] by [-5, 3] is a good choice. The maximum ener- gy, approximately 2.3807, occurs when r L 1.729. [0, 2] by [–1, 1]

154 Chapter 3 Exponential, Logistic, and Logarithmic Functions

76. 0 6 x 6 1.7115 (approx.) 78 5. = 0.65 = 65% 120 28 6. = 0.35 = 35% 80 7. 0.32x=48 gives x=150 8. 0.84x=176.4 gives x=210 9. 300 (1+0.05)=315 dollars [–1, 2] by [–2, 8] 10. 500 (1+0.45)=522.50 dollars 77. x …-20.0855 (approx.) Section 3.6 Exercises 1. Compound interest: A=1500(1+0.07)6≠$2251.10; simple interest: A=1500(1+0.07(6))≠$2130. 2. Compound interest: A=3200(1+0.08)4≠$4353.56; simple interest: A=3200(1+0.08(4))≠$4224. 3. Compound interest: A=12,000(1+0.075)7≠$19,908.59; simple interest: A=12,000(1+0.075(7))≠$18,300. [–40, 10] by [–1, 4] 4. Compound interest: A=15,500(1+0.095)12≠$46,057.58; x simple interest: A=15,500(1+0.095(12))≠$33,170. 78. log x - 2 log 3 7 0, so log(x/9) 7 0. Then 7 100 = 1, 9 0.07 20 7 5. A=15001 + ≠$2122.17 so x 9. a 4 b + + - 6 x 1 6 0.05 40 79. log(x 1) log 6 0, so log 0. 6. A=35001 + ≠$5752.67 6 a 4 b x + 1 Then 6 100 = 1, so x + 1 6 6, or x 6 5. The 0.038 240 6 7. A=40,5001 + ≠$86,496.26 a 12 b original equation also requires that x + 1 7 0, so the 300 solution is -1 6 6 5. 0.045 x 8. A=25,3001 + ≠$77,765.69 a 12 b (0.054)(6) ■ Section 3.6 Mathematics of Finance 9. A=1250e ≠$1728.31 10. A=3350e(0.062)(8)≠$5501.17 Exploration 1 11. A=21,000e(0.037)(10)≠$30,402.43 1. (0.044)(25) k A 12. A=8875e ≠$26,661.97 ` 0.07 24 10 1104.6 1 + - 1 ` a 4 b 20 1104.9 13. FV=500 # ≠$14,755.51 ` 0.07 30 1105 ` 4 40 1105 ` 0.06 48 50 1105.1 1 + - 1 ` # a 4 b 60 1105.1 14. FV=300 ≠$20,869.57 ` 0.06 70 1105.1 ` 4 80 1105.1 120 ` + 0.0525 - 90 1105.1 1 1 ` a 12 b 15. FV=450 # ≠$70,819.63 100 ` 1105.1 0.0525 A approaches a limit of about 1105.1. 12 2. y=1000e0.1≠1105.171 is an upper bound (and asymp- 0.065 300 1 + - 1 tote) for A(x). A(x) approaches, but never equals, this # a 12 b bound. 16. FV=610 ≠$456,790.28 0.065 Quick Review 3.6 12 - 1. 200 # 0.035 = 7 0.047 600.047 1 - 1 + 2. 150 # 0.025 = 3.75 a 12 b 12 17. PV=815.37 # ≠$43,523.31 1 # 3. 7.25% = 1.8125% - 4 0.065 3600.065 1 - 1 + 1 # a 12 b 12 4. # 6.5% L 0.5417% 18. PV=1856.82 ≠$293,769.01 12

Section 3.6 Mathematics of Finance 155

0.054 For #31–34, use the formula S=Pert. # (18,000) PV i a 12 b 31. Time to double: Solve 2=e0.09t, leading to 19. R=- = ≠$293.24 1 - (1 + i) n 0.054 -72 1 1 - 1 + t= ln 2≠7.7016 years. After 15 years: a 12 b 0.09 (0.09)(15) 0.072 S=12,500e ≠ $48,217.82. (154,000) 0.08t # a 12 b 32. Time to double: Solve 2=e , leading to PV i = 20. R=-n - ≠$1401.47 1 1 - (1 + i) 0.072 180 t= ln 2≠8.6643 years. After 15 years: 1 - 1 + 0.08 a 12 b S=32,500e(0.08)(15)≠ $107,903.80. In #21–24, the time must be rounded up to the end of the next 1 compounding period. 33. APR: Solve 2=e4r, leading to r= ln 2≠17.33%. 0.09 4t 83 4 21. Solve 2300 1 + = 4150:(1.0225)4t= ,so (0.1733)(15) a 4 b 46 After 15 years: S=9500e ≠$127,816.26 (using ln(83/46) the “exact” value of r). 1 L t= 6.63 years — round to 6 years 1 4 ln 1.0225 34. APR: Solve 2=e6r, leading to r= ln 2≠11.55%. 9 months (the next full compounding period). 6 After 15 years: S=16,800e(0.1155)(15)≠$95,035.15 (using 0.09 12t 22. Solve 8000 1 + = 16,000:(1.0075)12t=2, so the “exact” value of r). a 12 b In #35–41, the time must be rounded up to the end of the 1 ln 2 t= L 7.73 years — round to 7 years next compounding period (except in the case of continuous 12 ln 1.0075 compounding). 9 months (the next full compounding period). 4t + 0.04 = = 1 ln 2 L 0.08 12t 35. Solve 1 2: t 17.42, which 23. Solve 15,000 1 + = 45,000:(1.0067)12t=3, so a 4 b 4 ln 1.01 a 12 b rounds to 17 years 6 months. 1 ln 3 t= L 13.71 years — round to 13 years 0.08 4t 1 ln 2 12 ln 1.0067 36. Solve 1 + = 2: t = L 8.751, which a 4 b 4 ln 1.02 9 months (the next full compounding period). Note: A rounds to 9 years (almost by 8 years 9 months). graphical solution provides t L 13.78 years — round to 1 13 years 10 months. 37. Solve 1+0.07t=2:t = L 14.29, which rounds to 0.07 0.08 4t 24. Solve 1.5 1 + = 3.75:(1.02)4t=2.5, so 15 years. a 4 b ln 2 1 ln 2.5 t L t= L 11.57 years — round to 11 years 38. Solve 1.07 =2: t= 10.24, which rounds 4 ln 1.02 ln 1.07 9 months (the next full compounding period). to 11 years. r (365)(5) 0.07 4t 1 ln 2 25. Solve 22,000 1 + = 36,500: 39. Solve 1 + = 2: t = L 9.99, which a 365 b a 4 b 4 ln 1.0175 r 73 1/1825 rounds to 10 years. 1 + = , so r≠10.13%. 12t 365 a 44 b 0.07 1 ln 2 40. Solve 1 + = 2: t = a 12 b 12 ln(1 + 0.07/12) r (12)(5) 26. Solve 8500 1 + = 3 # 8500: ≠9.93, which rounds to 10 years. a 12 b 1 r 41. Solve 0.07t : ln years. 1 + = 31/60, so r≠22.17%. e =2 t= 2≠9.90 12 0.07 For #42–45, observe that the initial balance has no effect on 110 1/6 27. Solve 14.6 (1+r)6=22: 1+r= ,so the APY. a b 73 4 r≠7.07%. 1/8 0.06 25 42. APY= 1 + - 1 L 6.14% 28. Solve 18 (1+r)8=25: 1+r= , so r≠4.19%. a b a 18 b 4 0.0575 365 In #29 and 30, the time must be rounded up to the end of the 43. APY= 1 + - 1 L 5.92% next compounding period. a 365 b 0.063 0.0575 4t 1 ln 2 44. APY=e -1≠6.50% 29. Solve 1 + = 2: t= L 12.14 — a 4 b 4 ln 1.014375 0.047 12 45. APY= 1 + - 1 L 4.80% round to 12 years 3 months. a 12 b 0.0625 12t 0.05 12 30. Solve 1 + = 3: 46. The APYs are 1 + - 1 L 5.1162% and a 12 b a 12 b 1 ln 3 0.051 4 t= L 17.62 — round to 17 years 8 1 + - 1 L 5.1984%. So, the better investment 12 ln (1 + 0.0625/12) a 4 b months. is 5.1% compounded quarterly.

156 Chapter 3 Exponential, Logistic, and Logarithmic Functions

1 last payment will be less than $1050. A reasonable 47. The APYs are 5 % = 5.125% and e0.05-1≠5.1271%. 8 estimate of the final payment can be found by taking So, the better investment is 5% compounded continuously. the fractional part of the computed value of n above, (1 + i)n - 1 0.81, and multiplying by $1050, giving about $850.50. For #48–51, use the formula S=R . To figure the exact amount of the final payment, i -171 0.0726 1 - (1.01) 48. i= = 0.00605 and R=50, so solve 86,000=1050 +R (1.01)–172 12 0.01 (1.00605)(12)(25) - 1 S=50 L $42,211.46. (the present value of the first 171 payments, plus the 0.00605 present value of a payment of R dollars 172 months 0.155 from now). This gives a final payment of R≠$846.57. 49. i== 0.0129 » and R=50, so 12 (b) The total amount of the payments under the original (12)(20) (1.0129) - 1 plan is 360 # $884.61 = $318,459.60 . The total using S=50 L $80,367.73. # 0.0129 the higher# payments is 172 $1050 = $180,660 (or 0.124 171 $1050 + $846.57 = $180,396.57 if we use the 50. i= ; solve 12 correct amount of the final payment) — a difference 0.124 (12*30) of $137,859.60 (or $138,063.03 using the correct final 1 + --1 payment). a 12 b 250,000=R to obtain 57. (a) After 10 years, the remaining loan balance is 0.124 (1.01)120 - 1 12 86,000(1.01)120-884.61 L $80,338.75 R≠$239.42 per month (round up, since $239.41 will not 0.01 be adequate). (this is the future value of the initial loan balance, 0.045 minus the future value of the loan payments). With 51. i= = 0.00375; solve $1050 payments, the time required is found by solving 12 1 - (1.01)-n (1.00375)(12)(30) - 1 80,338.75=1050 ; this leads to 120,000=R to obtain R≠$158.03 0.01 0.00375 (1.01)–n≠0.23487, so n≠145.6 months, or about per month (round up, since $158.02 will not be adequate). 12.13 (additional) years. The mortgage will be paid off -n 1 - (1 + i) after a total of 22 years 2 months, with the final pay- For #52–55, use the formula A=R . i ment being less than $1050. A reasonable estimate of 0.0795 the final payment is (0.6)($1050)≠$630.00 (see the 52. i= = 0.006625; solve 12 previous problem); to figure the exact amount, solve - -(12)(4) - 145 1 - (1.006625) 1 (1.01) –146 9000=R to obtain R≠$219.51 80,338.75 = 1050 +R(1.01) , 0.006625 0.01 per month. which gives a final payment of R≠$626.93. 0.1025 (b) The original plan calls for a total of $318,459.60 in 53. i= = 0.0085417; solve # 12 payments; this plan calls for 120 $884.61+ - # # 1 - (1.0085417) (12)(3) 146 $1050=$259,453.20 (or 120 $884.61+ 4500=R to obtain R≠$145.74 # 0.0085417 145 $1050+$626.93=$259,030.13) — a savings per month (round up, since $145.73 will not be adequate). of $59,006.40 (or $59,429.47). 0.0875 58. One possible answer: The APY is the percentage increase 54. i= = 0.0072917; solve from the initial balance to the end-of-year balance 12 S(0) - S(1); specifically, it is S(1)/S(0)-1. Multiplying the ini- 1 - (1.0072917) (12)(30) 86,000=R to obtain R≠$676.57 tial balance by P results in the end-of-year balance being 0.0072917 multiplied by the same amount, so that the ratio remains per month (round up, since $676.56 will not be adequate). unchanged. Whether we start with a $1 investment, or a 0.0925 = r k 55. i= 0.0077083; solve 100,000= $1000 investment, APY= 1 + - 1. 12 a k b 1 - (1.0077083)-(12)(25) R to obtain R≠$856.39 per 59. One possible answer: The APR will be lower than the 0.0077083 APY (except under annual compounding), so the bank’s month (round up, since $856.38 will not be adequate). offer looks more attractive when the APR is given. 0.12 Assuming monthly compounding, the APY is about 56. (a) With i= = 0.01, solve 12 4.594%; quarterly and daily compounding give approxi- - 1 - (1.01) n mately 4.577% and 4.602%, respectively. 86,000=1050 ; this leads to 0.01 60. One possible answer: Some of these situations involve 860 19 counting things (e.g., populations), so that they can only (1.01)–n =1 - = , so n≠171.81 1050 105 take on whole number values — exponential models which months, or about 14.32 years. The mortgage will be predict, e.g., 439.72 fish, have to be interpreted in light of paid off after 172 months (14 years, 4 months). The this fact.

Section 3.6 Mathematics of Finance 157

Technically, bacterial growth, radioactive decay, and com- 63. False. The limit, with continuous compounding, is pounding interest also are “counting problems” — for exam- A = Pert = 100 e0.05 L $105.13. ple, we cannot have fractional bacteria, or fractional atoms 64. True. The calculation of interest paid involves compound- of radioactive material, or fractions of pennies. However, ing, and the compounding effect is greater for longer because these are generally very large numbers, it is easier to repayment periods. ignore the fractional parts. (This might also apply when one 65. A = P(1 + r/k)kt = 2250(1 + 0.07/4)4(6) L $3412.00. The is talking about, e.g., the population of the whole world.) answer is B. Another distinction: while we often use an exponential 12 model for all these situations, it generally fits better (over 66. Let x=APY.Then 1+x=(1+0.06/12) ≠1.0617. long periods of time) for radioactive decay than for most So x≠0.0617. The answer is C. of the others. Rates of growth in populations (esp. human 67. FV=R((1+i)n-1)/i= populations) tend to fluctuate more than exponential 300((1+0.00375)240-1)/0.00375≠$116,437.31. models suggest. Of course, an exponential model also fits The answer is E. well in compound interest situations where the interest 68. R=PV i/(1-(1+i)–n) rate is held constant, but there are many cases where =120,000(0.0725/12)/(1-(1+0.0725/12)–180) interest rates change over time. ≠$1095.44. 61. (a) Steve’s balance will always remain $1000, since The answer is A. interest is not added to it. Every year he receives 6% 69. The last payment will be $364.38. of that $1000 in interest: 6% in the first year, then 70. One possible answer: another 6% in the second year (for a total of The answer is (c). This graph shows the loan balance 2 # 6%=12%), then another 6% (totaling decreasing at a fairly steady rate over time. By contrast, 3 # 6%=18%), etc. After t years, he has earned 6t% of the early payments on a 30-year mortgage go mostly the $1000 investment, meaning that altogether he has toward interest, while the late payments go mostly toward 1000+1000 # 0.06t=1000(1+0.06t). paying down the debt. So the graph of loan balance (b) The table is shown below; the second column gives versus time for a 30-year mortgage at double the interest t values of 1000(1.06) . The effects of annual com- rate would start off nearly horizontal and more steeply pounding show up beginning in Year 2. decrease over time. Not 71. PV(10)=$81,109 ` ` Years Compounded Compounded ` ` PV(50)=$214,823 0 1000.00 1000.00 ` ` PV(100)=$254,050 1 1060.00 1060.00 ` ` 2 1120.00 1123.60 PV(150)=$249,303 ` ` 3 1180.00 1191.02 PV(200)=$249,902 ` ` 4 1240.00 1262.48 X = 420.45455 Y = 249999.98 ` ` PV(300)=$249,998 5 1300.00 1338.23 1 [0, 500] by [0, 300000] ` ` + -n = = 6 1360.00 1418.52 Since lim (1 i) lim 1 0, ` ` n: q n: q (1+i) 7 1420.00 1503.63 ` ` 1-(1+i)-n 8 1480.00 1593.85 = = R ` ` limq R . 9 1540.00 1689.48 n: i i ` ` 10 ` 1600.00 ` 1790.85 R 10,000 For the given annuity, = = $250,000. 62. (a) i 0.04 (1 + i)n - 1 72. (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=12 payments per year. [Ð10, 100] by [Ð100, 1400] (c) Each payment is R=$100. (b) The slope is P. r since that is the coefficient on t. (d) 20 years=20 # 12=240,so (c) The slope of the line increases as the interest rate 1 + 0.08 240 - 1 = A 12 B L increases and decreases as the interest rate decreases. F(240) 100 0.08 58,902.04 # # 12 = + = (d) PSimple (0.5) 300 300 0.03 0.5 $304.50; (e) = + 0.5 = PCompound (0.5) 300 (1 0.03) $304.47.

So PSimple (0.5) is greater. In the first half of the year, com-

pound interest grows more slowly than simple interest, so X = 240 simple interest will earn more money. By the end of the Y = 58902.042 year, compound interest grows more quickly that simple [0, 240] by [0, 70000] interest, so PCompound (1)=PSimple (1).

158 Chapter 3 Exponential, Logistic, and Logarithmic Functions

- 1 - (1 + i) n 7. f(x)=–2–3x-3 — starting from 2x, horizontally shrink 73. (a) Matching up with the formula A=R , 1 i by 3 , reflect across the y-axis, reflect across the x-axis, where i=r/k, with r being the rate and k being num- translate down 3 units. ber of payments per year, we find r=8%. (b) k=12 payments per year. (c) Each payment is R=$200. (d) 20 years=20 # 12=240,so 1 - 1 + 0.08 -240 = A 12 B L PV(240) 200 0.08 $23,910.86 [–2, 3] by [9, –1] 12 8. f(x)=2–3x+3 — starting from 2x, horizontally shrink by 1 , (e) 3 reflect across the y-axis, translate up 3 units.

X = 240 Y = 23910.858

[0, 240] by [0, 70000]

(f) Since interest is earned on the annuity’s remaining [–2, 3] by [–1, 9] balance each month, less money is required to achieve x 1 the future value of $48,000. Opinions will vary on 9. Starting from e , horizontally shrink by 2 , then translate 3 which option is better. right 2 units — or translate right 3 units, then horizontally 1 shrink by 2 . ■ Chapter 3 Review 1 1. f =-3 # 41/3 =-323 4 a 3 b 3 6 2 223 2. f - = 6 # 3-3/2 = = or a 2 b 227 23 3 For #3 and 4, recall that exponential functions have the form [–1, 3] by [0, 10] f(x)=a # bx. # 2 2 3. a = 3, so f(2) = 3 b = 6, b = 2, b = 22, 1 # x/2 10. Starting from ex, horizontally shrink by , then f(x)=3 2 3 = = # 3 = 3 = 1 = -1/3 4 4. a 2, so f(3) 2 b 1, b , b 2 , translate right units — or translate right 4 units, 2 3 = # -x/3 f(x) 2 2 1 –2x x then horizontally shrink by . 5. f(x)=2 +3 — starting from 2 , horizontally shrink 3 1 by , reflect across the y-axis, and translate up 3 units. 2

[–1, 3] by [0, 25] 100 [–4, 6] by [0, 10] 11. f(0) = = 12.5,,lim f(x) = 0 lim f(x) = 20 + : -q : q 1 5 3 x x 6. f(x)=2–2x — starting from 2x, horizontally shrink by , y-intercept: (0, 12.5). 2 reflect across the y-axis, reflect across x-axis. Asymptotes: y=0 and y=20 50 50 12. f(0) = = ,,lim f(x) = 0 lim f(x) = 10 5 + 2 7 x: -q x: q 50 y-intercept: 0, ≠(0, 7.14) a 7 b Asymptotes: y=0, y=10 13. It is an exponential decay function. [–2, 3] by [9, –1] lim f(x) = 2,lim f(x) = q x: q x: -q

Chapter 3 Review 159

Domain: (–q, q) Range: (0, 6) Continuous Increasing Symmetric about (1.20, 3) Bounded above by y=6 and below by y=0, the two asymptotes [–5, 10] by [–5, 15] No extrema lim f(x) = 6 , lim f(x) = 0 14. Exponential growth function x: q x: -q lim f(x) = q , lim f(x) = 1 x: q x: -q 18.

[–300, 500] by [0, 30] [–5, 5] by [–5, 15] Domain: (–q, q) 15. Range: (0, 25) Continuous Always increasing Symmetric about (–69.31, 12.5) Bounded above by y=25 and below by y=0, the two asymptotes No local extrema [Ð1, 4] by [Ð10, 30] = = limq g(x) 25 , limq g(x) 0 Domain: (–q, q) x: x: - Range: (1, q) For #19–22,# recall that exponential functions are of the form Continuous f(x) = a (1 + r)kx . Always decreasing 19. a=24, r=0.053, k=1; so f(x)=24 # 1.053x, where Not symmetric x=days. Bounded below by y=1, which is also the only asymptote 20. a=67,000, r=0.0167, k=1, so # x No local extrema f(x)=67,000 1.0167 , where x=years. = = q lim f(x) 1 , lim f(x) 1 # x/21 x: q x: -q 21. a=18, r=1, k= , so f(x)=18 2 , where 21 16. x=days. 1 1 22. a=117, r=- , k= , so 2 262 1x/262 f(x)=117 # = 117 # 2-x/262 , where x=hours. 2 For #23–26, recall that logistic functions are expressed in [–5, 5] by [–10, 50] c q q f(x)= . Domain: (– , ) + -bx q 1 ae Range: (–2, ) 30 Continuous 23. c=30, a=1.5, so f(2)= =20, + -2b Always increasing 1 1.5e 1 Not symmetric 30=20+30e–2b, 30e–2b=10, e–2b=, 3 Bounded below by y=–2, which is also the only 1 asymptote –2b ln e=ln L –1.0986, so b≠0.55. 3 No local extrema 30 lim g(x) = q , lim g(x) = -2 Thus,f(x) = . x: q x: -q 1 + 1.5e-0.55x 20 17. 24. c=20, a≠2.33, so f(3)= =15, 1 + 2.33e-3b 1 20=15+35e–3b, 35e–3b=5, e–3b=, 7 1 –3b ln e=ln L –1.9459, so b≠0.65. 7 20 [Ð5, 10] by [Ð2, 8] Thus,f(x) = . 1 + 2.33e-0.65x

160 Chapter 3 Exponential, Logistic, and Logarithmic Functions

20 25. c=20, a=3, so f(3)= =10, 1 + 3e-3b 10 1 20=10+30e–3b, 30e–3b=10, e–3b=,= 30 3 1 –3b ln e=ln L –1.0986, so b≠0.37. 3 20 Thus,f(x) L . [0, 10] by [–5, 5] 1 + 3e-0.37x 38. Translate left 1 unit, reflect across the x-axis, and translate 44 26. c=44, a=3, so f(5)= =22, up 4 units. 1 + 3e-5b 22 1 44=22+66e–5b, 66e–5b=22, e–5b=,= 66 3 1 –5b ln e=ln L –1.0986, so b≠0.22. 3 44 Thus,f(x) L . 1 + 3e-0.22x [–1.4, 17.4] by [–4.2, 8.2] 5 27. log2 32=log2 2 =5 log2 2=5 39. 4 28. log3 81=log3 3 =4 log3 3=4 1 1 29. log 23 10log 101= log 10= 3 3 3

1 - 7 =-7 =-7 30. ln =ln e 2 ln e 2e7 2 2 [Ð4.7, 4.7] by [Ð3.1, 3.1] 31. x=35=243 Domain: (0, q) 32. x=2y 1 Range: - , q L [–0.37, q) x c e b 33. = e-2 a y b Continuous y Decreasing on (0, 0.37]; increasing on [0.37, q) x = e2 Not symmetric y=xe2 Bounded below 1 1 a Local minimum at , - 34. = 10-3 a e e b a b b lim f(x) = q b x: q a = 1000 40. b=1000a 35. Translate left 4 units.

[Ð4.7, 4.7] by [Ð3.1, 3.1] Domain: (0, q) Range: [–0.18, q) Continuous by [–6, 7] [–6, 5] Decreasing on (0, 0.61]; increasing on [0.61, q) 36. Reflect across y-axis and translate right 4 units — or Not symmetric translate left 4 units, then reflect across the y-axis. Bounded below Local minimum at (0.61, –0.18) No asymptotes lim f(x) = q x: q 41.

[–6, 7] by [–6, 5] 37. Translate right 1 unit, reflect across the x-axis, and translate up 2 units. [Ð4.7, 4.7] by [Ð3.1, 3.1]

Chapter 3 Review 161

Domain: (–q, 0) ª (0, q) 1 the positive one,x = (240,009 - 1) L 99.5112 , works Range: [–0.18, q) 2 Discontinuous at x=0 in the original equation. q Decreasing on (– , –0.61], (0, 0.61]; 3x + 4 Increasing on [–0.61, 0), [0.61, q) 54. ln = 5 , so 3x+4=e5(2x+1). 2x + 1 Symmetric across y-axis 4 - e5 Bounded below Then x = L-0.4915 . 5 - Local minima at (–0.61, –0.18) 2e 3 and (0.61, –0.18) ln x 55. log x= No asymptotes 2 ln 2 lim f(x) = q , lim f(x) = q 56. log (6x2) = log 6 + log x2 = log 6 + 2 log |x| : q : -q 1/6 1/6 1/6 1/6 1/6 x x 2 ln |x| 2 ln |x| 2 ln |x| - + =- + =- - 42. = 1 1 - 1 ln 1/6 ln 6 1 ln 6 log x 57. log x= 5 log 5 3 = + 3 =- + 58. log1/2(4x ) log1/2 4 log1/2 x 2 3 log1/2 x =- - 2 3 log2x 3 log x =-2 - [0, 15] by [–4, 1] log 2 Domain: (0, q) 59. Increasing, intercept at (1, 0). The answer is (c). 60. Decreasing, intercept at (1, 0). The answer is (d). 1 61. Intercept at (–1, 0). The answer is (b). Range: - q, L (–q, 0.37] a e d 62. Intercept at (0, 1). The answer is (a). Continuous 63. A=450(1+0.046)3≠$515.00 Increasing on (0, e]L (0, 2.72], 0.062 (4)(17) 64. A=48001 + ≠$13,660.81 Decreasing [e, q)L [2.72, q) a 4 b Not symmetric 65. A=Pert Bounded above r kt 1 + - 1 1 L r a k b Local maximum at e, (2.72, 0.37) 66. i= , n=kt, so FV = R # a e b k r Asymptotes: y=0 and x=0 a k b lim f(x) = 0 x: q 0.055 (-12)(5) 550 1 - 1 + 43. x=log 4≠0.6021 a a 12 b b 67. PV = ≠$28,794.06 44. x=ln 0.25=–1.3863 0.055

ln 3 a b 45. x= ≠22.5171 12 ln 1.05 0.0725 (-26)(15) 5.4 953 1 - 1 + 46. x=e =221.4064 a a 26 b b –7 68. PV = ≠$226,396.22 47. x=10 =0.0000001 0.0725 ln 5 a 26 b 48. x=3+ ≠4.4650 ln 3 1 5 69. 20e–3k=50, so k=– ln L –0.3054. 2 3 2 49. log2x=2, so x=2 =4 3 7 70. 20e–k=30, so k=-ln L –0.4055. 50. log x= , so x=37/2=2723 ≠46.7654. 2 3 2 71. P(t)=2.041979 # 1.01296t, where x is the number of years # x x 2 # x 51. Multiply both sides by 23 , leaving (3 ) -1=10 3 , since 1900. In 2020, P(120)=2.041979 # 1.012960120≠9.57 x 2 # x x or (3 ) -10 3 -1=0. This is quadratic in 3 , leading million. 10 ; 2100 + 4 14.586176 to 3x= = 5 ; 226. Only 5+226 72. P(t)= , where x is the number of 2 1 + 2.043613e-0.024325t is positive, so the only answer is x=log3(5+226) years since 1900. In 2020, P(120)≠13.14 million. ≠2.1049. 73. (a) f(0)=90 units. 52. Multiply both sides by 4+e2x, leaving 50=44+11e2x, (b) f(2)≠32.8722 units. 1 6 so 11e2x=6. Then x= ln ≠–0.3031. (c) 2 11 53. log[(x+2)(x-1)]=4, so (x+2)(x-1)=104. The solutions to this quadratic equation are 1 x= (–1; 240,009 ), but of these two numbers, only 2 [0, 4] by [0, 90] Copyright © 2015 Pearson Education, Inc. ch03_p133_163.qxd 1/22/14 1:41 PM Page 162

162 Chapter 3 Exponential, Logistic, and Logarithmic Functions

74. (a) P(t)=123,000(1-0.024)t=123,000(0.976)t. log [H+]=–7.6 + –7.6 × –8 ln(90/123) [H ]=10 ≠2.51 10 (b) P(t)=90,000 when t= ≠12.86 years. ln 0.976 Milk of Magnesia: + 75. (a) P(t)=89,000(1-0.018)t=89,000(0.982)t. –log [H ]=10.5 log [H+]=–10.5 ln(50/89) + –10.5 × –11 (b) P(t)=50,000 when t= ≠31.74 years. [H ]=10 ≠3.16 10 ln 0.982 + - [H ] of Seawater 10 7.6 76. (a) P(0)≠5.3959 — 5 or 6 students. (b) = L 794.33 [H+] of Milk of Magnesia 10-10.5 (b) P(3)≠80.6824 — 80 or 81 students. (c) They differ by an order of magnitude of 2.9. (c) P(t)=100 when 1+e4–t=3, or t=4-ln 2 0.08 4t ≠3.3069 — sometime on the fourth day. 83. Solve 15001 + =3750: (1.02)4t=2.5, a b : q : 4 (d) As t , P(t) 300. 1 ln 2.5 # t so t= L 11.5678 years — round to 11 years 77. (a) P(t)=20 2 , where t is time in months. (Other 4 ln 1.02 # 12t # t/30 possible answers: 20 2 if t is in years, or 20 2 9 months (the next full compounding period). if t is in days). 84. Solve 12,500e0.09t=37,500: e0.09t=3, (b) P(12)=81,920 rabbits after 1 year. 1 × 19 so t= ln 3=12.2068 years. P(60)≠2.3058 10 rabbits after 5 years. 0.09 # t (c) Solve 20 2 =10,000 to find t=log2 500 700 85. t=133.83 ln L 137.7940 — about 11 years 6 months. ≠8.9658 months — 8 months and about 29 days. 250 # t t+2 78. (a) P(t)=4 2 =2 , where t is time in days. 500 86. t=133.83 ln L 308.1550 — about 25 years 9 months. (b) P(4)=64 guppies after 4 days. P(7)=512 guppies 50 after 1 week. 12 + 0.0825 # t 87. r=1 -1≠8.57% (c) Solve 42 =2000 to find t=log2 500=8.9658 days a 12 b — 8 days and about 23 hours. 88. r=e0.072-1≠7.47% 1 t/1.5 89. I=12 # 10(–0.0125)(25)=5.84 lumens 79. (a) S(t)=S # , where t is time in seconds. 0 a 2 b ln x –1 90. logb x= . This is a vertical stretch if e e. (c) If 1 g=S(60)=S # =S0 # , then 0 a 2 b a 2 b (There is also a reflection if 010. (There is also a reflection if 0

Chapter 3 Review 163

96. The model is T=75+145e–kt, and T(35)=150 3. 75 1 15 =75+145e–35k. Then e–35k=,so k=- ln 145 35 29 ≠0.0188. Finally, T=95 when 95=75+145e–kt, 1 20 so t=- ln L 105.17 minutes. k 145 (1 + i)n - 1 [Ð1, 6] by [0, 3] 97. (a) Matching up with the formula S=R , i 4. Each successive height will be predicted by multiplying where i=r/k, with r being the rate and k being the the previous height by the same percentage of rebound. number of payments per year, we find r=9%. The rebound height can therefore be predicted by the x (b) k=4 payments per year. equation y=HP where x is the bounce number. From the sample data, H=2.7188 and P≠0.778. (c) Each payment is R=$100. x # x - 5. y=HP becomes y≠2.7188 0.778 . 1 - (1 + i) n 98. (a) Matching up with the formula A=R , 6. The regression equation is 2.733 # 0.776x. Both H and i y≠ P are close to, though not identical with, the values in the where i=r/k, with r being the rate and k being the earlier equation. number of payments per year, we find r=11%. 7. A different ball could be dropped from the same original (b) k=4 payments per year. height, but subsequent maximum heights would in general (c) Each payment is R=$200. change because the rebound percentage changed. So P 99. (a) Grace’s balance will always remain $1000, since inter- would change in the equation. est is not added to it. Every year she receives 5% of 8. H would be changed by varying the height from which the that $1000 in interest; after t years, she has been paid ball was dropped. P would be changed by using a different 5t% of the $1000 investment, meaning that altogether type of ball or a different bouncing surface. she has 1000+1000 # 0.05t=1000(1+0.05t). 9. y=HP x (b) The table is shown below; the second column gives =H(eln P)x 0.05t values of 1000e . The effects of compounding con- =He(ln P)x tinuously show up immediately. =2.7188 e–0.251x x Not 10. ln y=ln (HP ) Years ` Compounded ` Compounded ` ` =ln H+x ln P 0 1000.00 1000.00 This is a linear equation. ` ` 1 1050.00 1051.27 ` ` 11. 2 1100.00 1105.17 Bounce Number ln (Height) ` ` 3 1150.00 1161.83 0 1.0002 ` ` 4 1200.00 1221.40 ` ` 1 0.76202 5 1250.00 1284.03 ` ` 2 0.50471 6 1300.00 1349.86 ` ` 7 1350.00 1419.07 3 0.23428 ` ` 8 1400.00 1491.82 4 –0.01705 ` ` 9 1450.00 1568.31 ` ` 5 –0.25125 10 ` 1500.00 ` 1648.72

Chapter 3 Project Answers are based on the sample data shown in the table. 2. Writing each maximum height as a (rounded) percentage of the previous maximum height produces the following [Ð1, 6] by [Ð1.25, 1.25] table. The linear regression produces Y=ln y≠–0.253x+1.005. Since ln y≠(ln P)x+ln H, the slope of the line is ln P Bounce Number Percentage Return and the Y-intercept (that is, the ln y-intercept) is ln H. 0 N/A 1 79% 2 77% 3 76% 4 78% 5 79% The average is 77.8%