Section 3.1 Exercises

Transcription

1 Section Exponential and Logistic Functions Chapter Exponential, Logistic, and Logarithmic Functions 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 Bounded below by y = 0, which is also the only asymptote lim f(x) = q, lim x: q f(x) = 0 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 Bounded below by y = 0, which is also the only asymptote lim g(x) = 0, lim g(x) = q x: q y = a b x y = x [, ] by [, 6] [, ] by [, 6] y = a b x y = x [, ] by [, 6] [, ] by [, 6] y = a 4 b x y =4 x [, ] by [, 6] [, ] by [, 6] y 4 = a 5 b x y 4 =5 x [, ] by [, 6] [, ] by [, 6] Copyright 05 Pearson Education, Inc

2 4 Chapter Exponential, Logistic, and Logarithmic Functions Exploration [ 4, 4] by [, 8] [ 4, 4] by [, 8] [ 4, 4] by [, 8] [ 4, 4] by [, 8] [ 4, 4] by [, 8] fx = x fx = x gx = e 04x fx = x gx = e 05x fx = x gx = e 06x fx = x gx = e 07x fx = x gx = e 08x , since ( 4) 5 = , since () 4 =95 Section Exercises Not an exponential function because the base is variable and the exponent is constant It is a monomial function Exponential function, with an initial value of and base of Exponential function, with an initial value of and base of 5 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 It is a power function 7 8 f(-) = 6 # - = 6 9 = 9 fa b = - # / = - 0 f(- ) = 8 # 4 -/ = f(x) = # a x b g(x) = # a x b 8 a 6 b 5 f(0) = # 5 0 = # = f(x) = # () x = # x/ 4 g(x) = # a x e b = e -x 8 ( ) / = 8 = 8 8 = 5 Translate f(x) = x by units to the right Alternatively, g(x) = x- = - # x = # x = # f(x), so it can be 8 8 obtained from f(x) using a vertical shrink by a factor of 8 k = 07 most closely matches the graph of f(x) k L 069 Quick Review -6 = -6 since (-6) = -6 5 = 5 B 8 since 5 = 5 and = 8 7 / =( ) / = = / =( ) 5/ = 5 = 5 [ 4, 4] by [, 8] Copyright 05 Pearson Education, Inc [, 7] by [, 8] 6 Translate f(x)= x by 4 units to the left Alternatively, g(x)= x±4 = # 4 x =8 # x =8 # f(x), so it can be obtained by vertically stretching f(x) by a factor of 8 [ 7, ] by [, 8]

3 Section Exponential and Logistic Functions 5 7 Reflect f(x) = 4 x over the y-axis 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 [, 9] 8 Reflect f(x) = x over the y-axis and then shift by 5 units to the right [, ] by [, 4] 4 Horizontally shrink f(x) = e x by a factor of, vertically stretch by a factor of, and shift down unit [, 7] by [ 5, 45] 9 Vertically stretch f(x) = 05 x by a factor of and then shift 4 units up [ 5, 5] by [, 8] 0 Vertically stretch f(x) = 06 x by a factor of and then horizontally shrink by a factor of [, ] by [, 4] Reflect f(x) = e x across the y-axis and horizontally shrink by a factor of [, ] by [, 8] 5 Graph (a) is the only graph shaped and positioned like the graph of y = b x, b 7 6 Graph (d) is the reflection of y = x across the y-axis 7 Graph (c) is the reflection of y = x across the x-axis 8 Graph (e) is the reflection of y = 05 x across the x-axis 9 Graph (b) is the graph of y = -x translated down units 0 Graph (f) is the graph of y = 5 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 Exponential decay: lim f(x) = 0; lim f(x) = q x: q 4 Exponential growth: lim f(x) = q; lim f(x) = 0 x: q 5 x x 7 0 [, ] by [ 0, ] [, ] by [, 5] Reflect f(x) = e x across the x-axis and y-axis Then, horizontally shrink by a factor of [ 05, 05] by [05, 5] [, ] by [ 5, 5] Copyright 05 Pearson Education, Inc

4 6 Chapter Exponential, Logistic, and Logarithmic Functions 7 x [ 05, 05] by [075, 5] 8 x 7 0 [ 05, 05] by [075, 5] 9 y = y, since x + 4 = (x + ) = ( ) x + = 9 x + 40 y = y, since # x = x = + x = x 4 y-intercept: (0, 4) Horizontal asymptotes: y=0, y = [ 0, 0] by [ 5, 5] 4 y-intercept: (0, ) Horizontal asymptotes: y = 0, y = 8 46 [, ] by [, 8] Domain: (- q, q) Range: (0, q) Always increasing Bounded below by y = 0, which is also the only asymptote lim f(x) = q, lim f(x) = 0 x: q [, ] by [, 8] Domain: (- q, q) Range: (0, q) Always decreasing Bounded below by y = 0, which is the only asymptote lim f(x) = 0, lim f(x) = q x: q 47 [ 5, 0] by [ 5, 0] 4 y-intercept: (0, 4) Horizontal asymptotes: y = 0, y = 6 [ 5, 0] by [ 5, 0] 44 y-intercept: (0, ) Horizontal asymptotes: y = 0, y = 9 [, ] by [, 9] Domain: (- q, q) Range: (0, q) Always increasing Bounded below by y = 0, which is the only asymptote lim f(x) = q, lim f(x) = 0 x: q [ 5, 0] by [ 5, 0] Copyright 05 Pearson Education, Inc

5 Section Exponential and Logistic Functions 7 48 Domain: (- q, q) Range: (0, q) Always decreasing Bounded below by y = 0, which is also the only asymptote lim f(x) = 0, lim f(x) = q x: q Domain: (- q, q) Range: (0, 5) Always increasing Symmetric about (069, 5) Bounded below by y = 0 and above by y = 5; both are asymptotes lim f(x) = 5, lim f(x) = 0 x: q Domain: (- q, q) Range: (0, 6) Always increasing Symmetric about (069, ) Bounded below by y = 0 and above by y = 6; both are asymptotes lim f(x) = 6, lim f(x) = 0 x: q For #5 and 5, refer to Example 7 on page 60 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() = 465,6 b = 80,6So, b = [, ] by [, 9] [, 4] by [, 7] [, 7] by [, 8] 80,6 B 465,6 L 074 Solving graphically, we find that the curve y = 465,6(074) t intersects the line y=800,000 at t L 00 Austin s population will pass 800,000 in 00 5 Let P(t) be Columbus s population t years after 990 Then with exponential growth, P(t) = P # 0 b t where P 0 =6,90 From Table 7, P()=6,90 b =7797,44 So, 797,44 b = B 6,90 L 0 Solving graphically, we find that the curve y = 6,90 (0) t intersects the line y=800,000 at t L Columbus s population will pass 800,000 in 0 5 Using the results from Exercises 5 and 5, we represent Austin s population as y=465,6(074) t and Columbus s population as y=6,90(0) t Solving graphically, we find that the curves intersect at t L 99 The two populations were equal, at 78,7, in 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() = 465,6 b = 80,6 So, 80,6 b = B 465,6 L 074 Solving graphically, we find that the curve y = 465,6(074) t intersects the line y=,000,000 at t L 88 Austin s population passed,000,000 in 08 Let P(t) be Columbus s population t years after 990 Then with exponential growth, P(t) = P 0 b t where P 0 =6,90 From Table 7, P()=6,90 b =7797,44 So, 797,44 b = B 6,90 L 0 Solving graphically, we find that the curve y = 6,90 (0) t intersects the line y=,000,000 at t L 444 Columbus s population will pass,000,000 in 0 From the results of above, Austin s population will reach,000,00 first, in Solving graphically, we find that the curve 79 y = intersects the line y=0 when + 40e -0009x t L 6967 Ohio s population stood at 0 million in (a) P(50) = L (50) e or,794,558 people 9875 (b) P(5) = L 9777 or (5) e 9,7,77 people (c) lim P(t) = 9875 or 9,875,000 people 57 (a) When t = 0, B = 00 (b) When t = 6, B L (a) When t = 0, C = 0 grams (b) When t = 0,400, C L 5647 After about 5700 years, 0 grams remain 59 False If a>0 and 0<b<, or if a<0 and b>, then f(x) = a # b x is decreasing c 60 True For f(x) = the horizontal asymptotes + a # b x are y=0 and y=c, where c is the limit of growth 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 05 Pearson Education, Inc

6 8 Chapter Exponential, Logistic, and Logarithmic Functions 6 For b>0, f(0)=b 0 = The answer is C 68 (a) f(x) decreases less rapidly as x increases 6 The growth factor of f(x) = a # b x is the base b The (b) y as x increases, g(x) decreases ever more rapidly answer is A 69 c = a : To the graph of ( a ) x apply a vertical stretch by b, 64 With x>0, a x >b x requires a>b (regardless of since f(ax + b) = ax + b = ax b = ( b )( a ) x whether x< or x>) The answer is B 70 a Z 0, c = 65 (a) 7 a 6 0, c = 7 a 7 0 and b 7, or a 6 0 and 0 6 b 6 7 a 7 0 and 0 6 b 6, or a 6 0 and b 7 74 Since 0<b<, lim ( + a # b x ) = q and x: -q c Thus, lim and [ 5, 5] by [, 5] lim ( + a # b x ) = x: - q + a # b x = 0 c Domain: (- q, q) lim Range: c- + a # b x = c e, q b (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 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, lim f(x) = q, lim x: -q f(x) = 0 [, ] by [ 7, 5] g(x) = 0, lim x: -q y = 0 g(x) = -q 66 (a) x = ( ) = 4, so x = 4 (b) x =, so x = (c) 8 x/ = 4 x +, ( ) x/ = ( ) x + # x = x +, x = 4x + 4, x = -4 (d) 9 x = x +, ( ) x = x +, x = x +, x = 67 (a) You have 4 =6 4 th great grandparents and they are in the 4 th generation (b) y= x with domain 0,,, etc (c) You have 6 =64 6 th great grandparents (d) You have 5 =,554,4 5 th great grandparents (e) It would take about 5 * 0 = 750 years to span 5 generations You would be directly related to about,554,4 or 84% of the world s 400,000,000 = 0084 population in 50 y Section Exponential and Logistic Modeling Quick Review 05 4% (07)() 4 (096)(5) 5 b = 60 = 4, so b = ;4 = ; 40 6 b = 9 4, so b = 9 B 4 = B 7 = 88 7 b= 0 6B b= 4 5B b= 06 4B b= 089 7B 7 Copyright 05 Pearson Education, Inc Section Exercises For # 0, use the model P(t) = P 0 ( + r) t r=009, so P(t) is an exponential growth function of 9% r=008, so P(t) is an exponential growth function of 8% r= 00, so f(x) is an exponential decay function of % 4 r= 000, so f(x) is an exponential decay function of 0% 5 r=, so g(t) is an exponential growth function of 00% 6 r= 095, so g(t) is an exponential decay function of 95% 7 f(x) = 5 # ( + 07) x = 5 # 7 x (x = years) 8 f(x) = 5 # ( + 00) x = 5 # 0 x (x = days) 9 f(x) = 6 # ( - 05) x = 6 # 05 x (x = months) 0 f(x) = 5 # ( ) = 5 # 0994 x (x = weeks) f(x) = 8,900 # ( - 006) x = 8,900 # 0974 x (x=years)

7 Section Exponential and Logistic Modeling 9 (x=years) 4 f(x) = 50,000 # ( + 007) x = 50,000 # 07 x f(x) = 8 # ( + 005) x = 8 # 05 x (x = weeks) f(x) = 5 # ( ) x = 5 # 0954 x (x = days) 5 f(x)= 06 # x/ (x=days) 6 f(x)= 50 # x/75 = 50 # x/5 (x = hours) 7 f(x)= 59 # (x=years) 8 f(x)= 7 # - x/6 - x/ (x=hours) 9 f 0 =, 875 = 5 = r +, so f(x)= # 5 x (Growth Model) 0 g 0 = -58, -464 = 08 = r +, so -58 g(x)= -58 # 08 x (Decay Model) For # and, use f(x) = f 0 # b x f Since f(5) = 4 # b 5 0 = 4, so f(x) = 4 # b x = 805, bfi= b= L 5 f(x) L 4 # 5 5B x 4, 4 f Since f(4) = # b 4 0 =, so f(x) = # b x = b = b= L 084 f(x) L # 084 4B x, 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 60 4 c=60, a=4, so f()= = 4, 60 = b, + 4b 60 96b=6, b=, thus f(x)= 8 + 4a x 8 b 8 5 c=8, a=7, so f(5)=, + 7b = =+4bfi, 4bfi=96, bfi=, b=, thus f(x) L 5B 4 L # 0844 x 0 6 c=0, a=5, so f()= = 5, 0 = b, + 5b 5 75b =5, b = b=, 75 = 5, B 5 L thus f(x) L + 5 # 0585 x 0 7 c=0, a=, so f()= + b = 0, 0 = 0 + 0b, 0b =0, b = b=,, B L thus f(x)= + # 058 x 60 8 c=60, a=, so f(8)= + b 8 = 0, 60 = b8, 90b =0, b = b=,, 8B L thus f(x)= + # 087 x 9 P(t) = 76,000(049) t ; P(t)=,000,000 when t 07 years, or in the year 00 0 P(t) = 478,000(068) t ; P(t)=,000,000 when t years, or in the year 0 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 7665 years after 890 in 966 The model is P(t) = 400(05) t (a) In 90: about P(0) 6554 In 945: about P(5) 95 (b) P(t)=0,000 when t 704 years after 90: about 980 (a) y = 66a t/4, where t is time in days b (b) After 8 days 4 (a) y = 5a t/65, where t is time in days b (b) After 748 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 6 One possible answer: Exponential functions and logistic functions are similar in the sense that they are always increasing or always decreasing They differ, however, in the sense that logistic functions have both an upper and a lower limit to their growth (or decay), while exponential functions generally have only a lower limit (Exponential functions just keep growing) 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 8 One possible answer: The number of atoms of a radioactive substance that change to a nonradioactive state in a given time is a fixed percentage of the number of radioactive atoms initially present So the time it takes for half of the atoms to change state (the half-life) does not depend on the initial amount 9 When t=, B 00 the population doubles every hour 40 The half-life is about 5700 years Copyright 05 Pearson Education, Inc

8 40 Chapter Exponential, Logistic, and Logarithmic Functions For #4 and 4, use the formula P(h)=47 is miles above sea level # 4 P(0)= /6 =4 lb/in 05 h/6, where h 4 P(h) = 47 # 05 h/6 intersects y = 5 when h 90 miles above sea level #,0,64 48 P(t) L which is the same model as + 60 e t, the solution in Example 8 of Section Note that t represents the number of years since 900 [, 9] by [, 9] 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 0 is P() L 4845, or 4,84,500 people This is an overestimate of 565,000 people, 565,000 an error of L 05 = 5%,80, The exponential regression model using data is P(t) = 08400(04465) t, where P(t) is measured in thousands of people and t is years since 900 The predicted population for Phoenix for 0 is P() L 6586, or,658,600 people This is an overestimate of,89,00 people,,89,00 an error of L 08 = 8%,469,4000 The equations in #45 and 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 690 about 7 days 46 (a) P(0)= (b) P(t)=600 when t 45 after 4 or 5 years (c) As t : q, P(t) : 00 the population never rises above this level 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) L 9 or about 9 million people [ 0, 0] by [ 0, 400] [0,0] by [0,500000] 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 [0, 00] by [0, 0] P(x) L, where x is the number of e -0044x years since 800 and P is measured in millions As x : q, P(x) : 564, or nearly 6 million, which is significantly less than New York s population limit of 0 million The population of Arizona, according to our models, will not surpass the population of New York Our graph confirms this [0, 500] by [0, 5] 5 False This is true for logistic growth, not for exponential growth 5 False When r<0, the base of the function, +r, is merely less than 5 The base is 049=+0049, so the constant percentage growth rate is 0049=49% The answer is C 54 The base is 084=-066, so the constant percentage decay rate is 066=66% The answer is B 55 The growth can be modeled as P(t)= # t/4 Solve P(t)=000 to find t 986 The answer is D 56 Check S(0), S(), S(4), S(6), and S(8) The answer is E (a) P(x) L where x is the number of + 790e -007x, years since 900 and P is measured in millions P() 089, or about 08,900,000 people (b) The logistic model underestimates the 000 population by about 7 million, an error of around 08% (c) The logistic model predicted a value closer to the actual value than the exponential model, perhaps indicating a better fit Copyright 05 Pearson Education, Inc

9 Section Logarithmic Functions and Their Graphs 4 58 (a) Using the exponential growth model and the data from 900 0, Mexico s population can be represented by M(x) L 59 * 0 x, where x is the number of years since 900 and M is measured in milions Using the data for the United States, and the exponential growth model, the population of the United States can be represented by P(x) L 87 * 0 x,where x is the number of years since 900 and P is measured in millions Since Mexico s rate of growth outpaces the United States rate of growth, the model predicts that Mexico will eventually have a larger population Our graph indicates that this will occur at x L, or 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, 80] by [ 00, 000] (b) Using logistic growth models and the same data, 5968 M(x) = + 4e -0049x P(x) = e -006x Using this model, Mexico s population will not exceed that of the United States, confirmed by our graph Exploration [ 6, 6] by [ 4, 4] Same graph as part (c) According to the logistic growth models, the maximum sustainable populations are: Mexico 60 million people United States 758 million people (d) Answers will vary However, a logistic model acknowledges that there is a limit to how much a country s population can grow 59 sinh(-x) = e-x - e -(-x) = e-x - e x = -a ex - e -x b = -sinh(x), so the function is odd 60 cosh(-x) = e-x + e -(-x) = e-x + e x =cosh(x), so the function is even 6 (a) [ 0, 500] by [ 00, 000] = ex + e -x e x - e -x sinh(x) cosh (x) = e x + e -x e x - e -x = # = ex - e -x e x -x = tanh(x) e x + e -x + e Quick Review 5 = = = 0 = = = 9 8 = 5 = 7 5 / 8 0 / 9 0 a / a / e b = e -/ e b = e -/ Section Exercises log 4 4= because 4 =4 log 6 =0 because 6 0 = log =5 because 5 = 4 log 8=4 because 4 =8 5 log 5 5 = because 5 / = 5 6 log 6 = - because 6 -/5 = = /5 5 6 Copyright 05 Pearson Education, Inc

10 4 Chapter Exponential, Logistic, and Logarithmic Functions 7 log 0 = 8 log 0,000=log 0 4 =4 9 log 00,000=log 0 5 =5 0 log 0 4 = 4 log =log 0 / 0 = log =log 0 / = ln e = 4 ln e 4 = 4 5 ln =ln e = e 6 ln =ln e 0 =0 7 ln 4 e=ln e /4 = 4 8 ln e = ln 7 7 e-7/ = - 9, because b logb = for any b>0 0 8, because b logb8 = 8 for any b>0 0 log (05) = 0 log 0 (05) = 05 0 log 4 = 0 log 0 4 = 4 e ln 6 = e loge 6 =6 4 e ln (/5) = e loge(/5) = /5 5 log and log and log ( 4) is undefined because 4<0 8 log ( 54) is undefined because 54<0 9 ln and e ln 07 0 and e 0 07 ln ( 049) is undefined because 049<0 ln ( ) is undefined because <0 x=0 =00 4 x=0 4 =0,000 5 x=0 = =0 0 6 x=0 = = f(x) is undefined for x> The answer is (d) 8 f(x) is undefined for x< The answer is (b) 9 f(x) is undefined for x< The answer is (a) 40 f(x) is undefined for x>4 The answer is (c) 4 Starting from y=ln x: translate left units 4 Starting from y=ln x: translate up units [ 5, 5] by [, 4] 4 Starting from y=ln x: reflect across the y-axis and translate up units [ 4, ] by [, 5] 44 Starting from y=ln x: reflect across the y-axis and translate down units [ 4, ] by [ 5, ] 45 Starting from y=ln x: reflect across the y-axis and translate right units [ 7, ] by [, ] 46 Starting from y=ln x: reflect across the y-axis and translate right 5 units [ 6, 6] by [ 4, 4] 47 Starting from y=log x: translate down unit [ 5, 5] by [, ] [ 5, 5] by [, ] Copyright 05 Pearson Education, Inc

11 Section Logarithmic Functions and Their Graphs 4 48 Starting from y=log x: translate right units [ 5, 5] by [, ] 49 Starting from y=log x: reflect across both axes and vertically stretch by 54 Domain: (, q) Range: ( q, q) Always increasing Not bounded Asymptote at x= lim f(x)=q [ 8, ] by [, ] 50 Starting from y=log x: reflect across both axes and vertically stretch by 55 Domain: (, q) Range: ( q, q) Always increasing Not bounded Asymptote: x= lim [, 8] by [, ] f(x)=q [ 8, 7] 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, ] 5 Starting from y=log x: reflect across both axes, translate right unit, vertically stretch by, translate up unit 56 Domain: (, q) Range: ( q, q) Always decreasing Not bounded Asymptote: x= lim [, 8] by [, ] f(x) = -q 5 [ 6, ] by [, ] [, 9] by [, ] Domain: (, q) Range: ( q, q) Always decreasing Not bounded Asymptote: x= lim [, 7] by [, ] f(x)= q Copyright 05 Pearson Education, Inc

12 44 Chapter Exponential, Logistic, and Logarithmic Functions 57 6 (a) The exponential regression model is x, where x is the year and y is the population (b) # [, 7] by [, ] Domain: (0, q) Range: ( q, q) Increasing on its domain No symmetry Not bounded Asymptote at x=0 lim f(x) = q 58 [ 7,, ] by [ 0, 0, ] Domain: ( q,) Range: ( q, q) Decreasing on its domain No symmetry Not bounded Asymptote at x= lim f(x) = q 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) 60 I= # lumens (a) A magnitude earthquake is times more 00 = 0 powerful than a magnitude earthquake A magnitude 5 earthquake is = 00 times more pow- 00, erful than a magnitude earthquake (b) (c) Solving graphically, we find that the curve y= # x intersects the line y=500,000 at t=047 (d) Not in most cases as populations will not continue to grow without bound 6 True, by the definition of a logarithmic function 64 True, by the definition of common logarithm 65 log 000 The answer is C 66 log but 5 log 075 The answer is A 67 The graph of f(x)=ln x lies entirely to the right of the origin The answer is B 68 For f(x) = # x, f - (x) = log (x/) because f - (f(x)) = log ( # x /) =log x =x The answer is A 69 f(x) ` x ` log x 70 [900, 00] by [00000, ] Domain ` ( q, q) ` (0, q) Range ` (0, q) ` ( q, q) Intercepts ` (0, ) ` (, 0) Asymptotes ` y=0 ` x=0 [ 6, 6] by [ 4, 4] f(x) ` 5 x ` log 5 x Domain ` ( q, q) ` (0, q) Range ` (0, q) ` ( q, q) [0, 6] by [0, 0000] (c) Ground motion = 0 x where x is the magnitude of the earthquake, so y =0 x (d) y=0 x, so x =log y (e) Extremely large values can be represented by much smaller values (f) Yes Intercepts ` (0, ) ` (, 0) Asymptotes ` y=0 ` x=0 [ 6, 6] by [ 4, 4] Copyright 05 Pearson Education, Inc

13 Section 4 Properties of Logarithmic Functions 45 e 7 b= e The point that is common to both graphs is (e, e) 4 log 0 = x 5 y - 5 x 5 y ( 4) =x y x y -4 = [ 7, 57] by [, 4] 7 0 is not in the domain of the logarithm functions because 0 is not in the range of exponential functions; that is, a x is never equal to 0 7 Reflect across the x-axis 74 Reflect across the x-axis Section 4 Properties of Logarithmic Functions Exploration log ( # 4) 09009, log +log log a , log 8-log b log 09009, log (000) log 5=log a 0 =log 0-log -000 b = log 6=log 4 =4 log 04 log =log 5 =5 log 5055 log 64=log 6 =6 log log 5=log 5 = log 5= log a 0 b =(log 0-log ) 9794 log 40 = log (4 # 0) = log 4 + log 0 L 6006 log 50 = log a 00 b = log 00 - log L The list consists of,, 4, 5, 8, 6, 0, 5,, 40, 50, 64, and 80 Exploration False False; log (7x)=log 7+log x True 4 True x 5 False; log =log x-log True 7 False; log 5 x =log 5 x+log 5 x= log 5 x 8 True Quick Review 4 log 0 = ln e = ln e = u - v 7 6 u - v = v 7- v5 = - - (-) u u x 7 (x 6 y ) / =(x 6 ) / (y ) / = y y 9 8 (x 8 y ) /4 =(x 8 ) /4 (y ) /4 = (u v -4 ) / - u v 9 = (7 u 6 v - 6 / ) u v - = u (x - y ) - 0 (x y - ) - = x4 y -6 x -9 y = x 6 y Section 4 Exercises ln 8x=ln 8+ln x= ln +ln x ln 9y=ln 9+ln y= ln +ln y log =log -log x x 4 log =log -log y y 5 log y 5 =5 log y 6 log x = log x 7 log x y =log x +log y = log x+ log y 8 log xy =log x+log y =log x+ log y x 9 ln ln x -ln y = lnx- ln y y = 0 log 000x 4 =log 000+log x 4 =+4 log x x log = (log x-log y)= 4 log x - 4B y 4 4 log y x ln = (ln x-ln y)= ln x - y ln y log x+log y=log xy 4 log x+log 5=log 5x 5 ln y-ln =ln(y/) 6 ln x-ln y=ln(x/y) 7 log x=log x / =log x 8 log z=log z /5 =log 5 z 5 9 ln x+ ln y=ln x +ln y =ln (x y ) 0 4 log y-log z=log y 4 -log z=log a y4 z b 4 log (xy)- log (yz)=log (x 4 y 4 )-log (y z ) =log a x4 y 4 y z b = log a x4 y z b ln (x y)+ ln (yz )=ln (x 9 y )+ln (y z 4 ) =ln (x 9 y 5 z 4 ) x 6 Copyright 05 Pearson Education, Inc

14 46 Chapter Exponential, Logistic, and Logarithmic Functions In # 8, natural logarithms are shown, but common (base-0) logarithms would produce the same results ln 7 ln L 8074 ln 9 4 ln 5 L 895 ln 75 5 L 487 ln 8 ln 59 6 ln L 6 ln 7 = -ln ln 05 ln L ln = -ln ln 0 ln 5 L -09 ln x 9 log x= ln ln x 0 log 7 x= ln 7 ln(a + b) log (a+b)= ln 40 Starting from g(x)=ln x: vertically shrink by a factor /ln 7 05 [, 0] by [, ] 4 Starting from g(x)=ln x: reflect across the x-axis, then vertically shrink by a factor /ln 09 [, 0] by [, ] 4 Starting from g(x)=ln x: reflect across the x-axis, then shrink vertically by a factor of /ln 5 L 06 ln(c - d) log 5 (c-d)= ln 5 log x log x= log log x 4 log 4 x= log 4 log(x + y) 5 log / (x+y)= log (/) log(x + y) = - log log(x - y) log(x - y) 6 log / (x-y)= = - log(/) log 7 Let x=log b R and y=log b S Then b x =R and b y =S, so that R S = bx b y = bx - y [, 0] by [, ] 4 (b): [ 5, 5] by [, ], with Xscl= and Yscl= (graph y=ln(-x)/ ln 4) 44 (c): [, 8] by [, ], with Xscl= and Yscl= (graph y=ln(x-)/ ln 6) 45 (d): [, 8] by [, ], with Xscl= and Yscl= (graph y=ln(x-)/ ln 05) 46 (a): [ 8, 4] by [ 8, 8], with Xscl= and Yscl= (graph y=ln(-x)/ ln 07) 47 log b a R S b = log b b x - y = x - y = log b R - log b S 8 Let x=log b R Then b x =R, so that R c =(b x ) c = b c # x log b R c = log b b c # x = c # x = c log b R 9 Starting from g(x)=ln x: vertically shrink by a factor /ln 4 07 [, 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 () [, 0] by [, ] Copyright 05 Pearson Education, Inc

15 Section 4 Properties of Logarithmic Functions [, 9] by [ 5, ] Domain: (0, q) Range: ( q, q) Always decreasing Asymptote: x=0 lim f(x)= q ln (9x) f(x)= (9x)= 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 f(x)=q, Domain: (0, q) Range: ( q, q) Always increasing Asymptote: x=0 lim log / [ 0, 0] by [, ] [, 9] by [, 8] f(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 ) 50 5 (a) R=log + 45 = log (b) R=log 00 4 lim x: q ln a b + 5 = log I 5 log 0005(40)= 0094, so = I= # L lumens I 54 log = 005(0)= 05,so I= # 0-05 L lumens 55 From the change-of-base formula, we know that f(x) = log x = ln x 090 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 From the change-of-base formula, we know that log x f(x)=log 08 x= log 08 = # log x L-0 log x log 08 f(x) can be obtained from g(x)=log x by reflecting across the x-axis and vertically stretching by a factor of approximately 0 57 True This is the product rule for logarithms 58 False The logarithm of a positive number less than is negative For example, log 00= 59 log =log ( # 4)=log +log 4 by the product rule The answer is B 60 log 9 64=(ln 64)/(ln 9) by the change-of-base formula The answer is C 6 ln x 5 =5 ln x by the power rule The answer is A 6 log / x = log / x ln x = ln(/) ln x = ln - ln = - ln x ln = - log x The answer is E 6 (a) f(x)=75 # x 50 (b) f(7) 49,66 (c) ln(x) p ln(y) [0, ] by [0, 5] (d) ln(y)=500 ln x+0 (e) a 5, b so f(x)=e x 5 =ex 5 7x 5 The two equations are the same 64 (a) f(x)=8095 # x 0 (b) f(9)=8095 (9) 0 60 # Copyright 05 Pearson Education, Inc

16 48 Chapter Exponential, Logistic, and Logarithmic Functions (c) ln(x) p ln(y) [05, 5] by [8, ] (d) ln(y)= 0 ln (x)+09 (e) a 0, b, so f(x)=e # x 0 809x 0 65 (a) log(w) p log(r) [, ] by [6, 8] (b) log r=( 00) log w+6 (c) [, ] by [6, 8] (d) log r=( 00) log (450)+6 58, r 769, 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 66 log4 = log log5 = log(0/) = log0 - log = - log log6 = log + log log8 = log( ) = log log9 = log( ) = log log = log + log4 = log + log log5 = loga 0 * b = log0 - log + log = - log + log log6 = log( 4 ) = 4log log8 = log + log9 = log + log log0 = log(0 * ) = log0 + log = + log log4 = log( * 8) = log + log log5 = loga 00 = log00 - log4 = - log 4 b log7 = log( ) = log log0 = log(0 * ) = log0 + log = + log log = log( 5 ) = 5log log6 = log4 + log9 = log + log log40 = log(0 * 4) = log0 + log4 = + log log45 = log5 + log9 = - log + log log48 = log + log6 = log + 4log log50 = loga 00 = log00 - log = - log b log54 = log + log7 = log + log log60 = log(0 * 6) = log0 + log6 = + log + log log64 = log( 6 ) = 6log log7 = log8 + log9 = log + log log75 = loga 00 * b 4 log80 = log(0 * 8) = log0 + log8 = + log log8 = log( 4 ) = 4log Copyright 05 Pearson Education, Inc log90 = log(0 * 9) = log0 + log9 = + log log96 = log + log = log + 5log 67 Since =458 /5 and log (458 / )= log 458, 5 we need to find log 458 L (using tables, before calculators were available) and then find L, so L L 540 For #68 and 69, solve graphically x x (a) [, 9] by [, 8] Domain of f and g: (, q) (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 + log

17 Section 5 Equation Solving and Modeling 49 7 Recall that y=log a x can be written as x=a y Let y=log a b a y =b log a y =log b y log a =log b log b y= log a b log a = 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 04 log x log e Thus, y is a constant function 7 [, 9] by [, ] Domain: (, q) Range: ( q, q) Increasing Vertical asymptote: x= lim f(x)=q One-to one, hence invertible (f (x)= e ex ) 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 4 4 # 0 0 is nine orders of magnitude greater than 4 # 0 Quick Review 5 In # 4, graphical support (ie, 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)) = 0 (log x )/ = 0 log x = x and g(f(x))= log(0 x/ ) = log(0 x ) = x f(g(x)) = ln(ex ) = (x) = x and (/ ln x) g(f(x))= e = e ln x = x 4 f(g(x)) = log(0 x/6 ) = 6 log(0 x/6 ) = 6(x/6) = x and g(f(x)) = 0 ( log x )/6 = 0 (6 log x)/6 = 0 log x = x * 0 8 km 6 * 0-5 m 7 60,000,000,000,000,000,000, (6 zeros between the decimal point and the ) 9 (86 * 0 5 )( * 0 7 ) = (86)() * = 5766 * 0 8 * * 0 = * (-6) = 6 * 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 a x/ 4 b = a x/ 4 b = 6 a x/ 4 b = a 4 b x = x = 6 # 5 x/4 = 50 5 x/4 = 5 5 x/4 = 5 x 4 = x = 4 # 4 x/ = 96 4 x/ = 4 x/ = x = 5 4 5/ x = x/ = 0, so -x/ =, and therefore x = x/4 = 5, so -x/4 =, and therefore x = -4 7 x = 0 4 = 0,000 8 x= 5 = 9 x - 5 = 4 -, so x = = 55 Copyright 05 Pearson Education, Inc

18 50 Chapter Exponential, Logistic, and Logarithmic Functions 0 - x = 4, so x = - 0 Multiply both sides by # ln 4 or ( x ) - 6 # x, leaving ( x ) + = 6 # x, x + = 0 This is quadratic in x, leading to x = ln 06 = log 06 4 L 45 ln( ; ) x = 6 ; 6-4 = ; Then x = ln 6 x = ln 098 = log ln L -644 = log ( ; ) L ;54 e 005x = 4, so 005x = ln 4, and therefore Multiply both sides by e x, leaving (e x ) + = 8e x, or This is quadratic in leading to x = (e ln 4 L x ) - 8e x + = 0 e x, e 0045x Then =, so 0045x = ln, and therefore e x = 8 ; 64-4 = 4 ; 5 x = ln L 446 x=ln(4 ; 5) L ; e -x =, so -x = ln This is quadratic in e x, leading to, and therefore -5 ; e x = = -5 ; 7 Of these two x = -ln 4 4 L numbers, only = is positive, so x = ln 6 e -x = 5, so -x = ln 5 4, and therefore 069 x = -ln L and therefore 00 = + 5e0x, so e 0x = 50 = 006, x = ln 006 L ln(x-)= and therefore, so x - = e/, x = + e / L log(x + ) = -, so x + = 0 -, and therefore x = = We must have x(x + ) 7 0, so x 6 - or x 7 0 Domain: (- q, -) h (0, q) ; graph (e) 0 We must have x 7 0 and x + 7 0, so x 7 0 Domain: (0, q); graph (f) 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 and x + 7 0, so x 7 0 Domain: (0, q); graph (c) We must have x 7 0 Domain: (0, q); graph (a) 4 We must have x 7 0, so x Z 0 Domain: (- q, 0) h (0, q) ; graph (b) 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 e, leaving e ln x = e 4, or x = e 4 Then x = e L 789 or x = -e L Write both sides as powers of 0, leaving 0 log x4 = 0, or x 4 = 00 Then x = 0, and x = ;0 8 Write both sides as powers of e, leaving e ln 6 x = e, or x 6 = e Then x = e 4, and x = ; e 9 Multiply both sides by # = # x, or ( x ) - # x, leaving ( x ) - x - = 0 This is quadratic in x, leading to x ; = = 6 ; 7 Only is positive, so the only answer is ln(6 + 7) x = = log ln (6 + 7) L and therefore 50 = + 95e-06x, so e -06x = 57, x = -06 ln 57 L Multiply by, then combine the logarithms to obtain x + ln so x + = x x = 0 Then x + x = e 0 =, The solutions to this quadratic equation are x = 6 Multiply by, then combine the logarithms to obtain x x log Then so x + 4 = 0 = 00, x + 4 = x = 00(x + 4) The solutions to this quadratic equation 00 ; are x = = 50 ; 09 The original equation requires that x 7 0, so is extraneous; the only actual solution is x = L 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) 8 log[(x-)(x+5)]= log, so (x-)(x+5)=9, or x + x - 9 = 0 Then x = ; + Copyright 05 Pearson Education, Inc - ; = ; L 08 = - The actual ; 85 solution is x = - since x- must + 85 L 098; be positive, the other algebraic solution, x= - is extraneous - 85, 9 A $00 bill has the value of 000, or 0, dimes so they differ by an order of magnitude of

19 Section 5 Equation Solving and Modeling 5 40 A -kg hen weighs 000, or # 0 grams while a 0-g canary weighs #, 48 (a) Stomach acid: log [H + ]=0 0 grams They differ by an order of log [H + ]= 0 magnitude of [H + ]=0 0 =* =5 They differ by an order of magnitude of 5 Blood: log [H + ]=74 log [H + ]= =8 They differ by an order of magnitude [H + ]= *0 8 of 8 [H + ] of stomach acid 4 Given (b) = 0- L 5 * b = 0 log I [H ] of blood 0 = 95 I (c) They differ by an order of magnitude of 54 0 b = 0 log I The equations in #49 and 50 can be solved either algebraically = 65, I or graphically; the latter approach is generally faster 0 we seek the logarithm of the ratio I /I 49 Substituting known information into 0 log I - 0 log I T(t)= T leaves T(t)=+70e kt m + (T 0 - T m )e -kt = b I 0 I - b 0 0 alog I - log I Using T()=50=+70e k, we have e k = b = I 0 I 0 k= - Solving T(t)=0 yields 0 log I ln 5, so 5 L = 0 t 84 minutes I log I 50 Substituting known information into T(t) = = T leaves T(t)=65+85e kt m + (T 0 - T m )e -kt I Using T(0)=0=65+85e 0k, we have The two intensities differ by orders of magnitude e 44 Given 0k = so k= ln 008 Solving 57, 0 57 b = 0 log I T(t)=90 yields t 959 minutes = 70 I 0 5 (a) b = 0 log I = 0, I 0 we seek the logarithm of the ratio I /I 0 log I I 0-0 log I I 0 = b - b 0 alog I I 0 - log I I 0 b = log I I = 60 log I = 6 I The two intensities differ by 6 orders of magnitude 45 Assuming that T and B are the same for the two quakes, we have 79 = log a - log T + B and 66= log a - log T + B, so = =log(a /a ) Then a /a = 0, so a L 995a the Mexico City amplitude was about 0 times greater 46 If T and B were the same, we have 7 = log a - log T + B and 66 = log a - log T + B, so 7-66=06= log(a Then a /a = 0 06 /a ), so a L 98a Kobe s amplitude was about 4 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 (b) (c) T(0)+0= C 5 (a) (b) Copyright 05 Pearson Education, Inc [0, 40] by [0, 80] [0, 40] by [0, 80] T(x) L 7947 # 09 x [0, 5] by [0, 90] [0, 5] by [0, 90] T(x) L 7996 # 09 x (c) T(0) =7996 C

20 5 Chapter Exponential, Logistic, and Logarithmic Functions 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 54 Answers will vary 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] 56 Exponential the scatterplot of (x, y) is exactly exponential (The data can be modeled by y= # x ) [0, 5] by [0, 00] 57 Exponential the scatterplot of (x, y) is exactly exponential (The data can be modeled by y= # x ) 6 x = x = 5 x-=5 x= The answer is B 6 ln x= e ln x =e x = e 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 log a log a T - log a = 8-6 T = a a = 0 =00 a The answer is E 64 As the second term on the right side of the formula T(t) = T m + (T 0 - T m )e -kt indicates, and as the graph confirms, the model is exponential The answer is A A logistic regression af(x) = most + 596e b x closely matches the data, and would provide a natural cap to the population growth at approx 945,000 people (Note: x=number of years since 900) The logistic regression model af(x) = + 58e -0097xb matches the data well and provides a natural cap of 0 million people (Note: x=number of years since 900) 67 (a) [0, 5] by [0, 0] 58 Linear the scatterplot of (x, y) is exactly linear (y=x+) [, ] by [0, 0] 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, ] 59 False The order of magnitude of a positive number is its common logarithm 60 True In the formula T(t) = T m + (T 0 - T m )e -kt, the term (T 0 - T m )e -kt goes to zero as t gets large, so that T(t) approaches T m [, ] by [0, ] As c increases, the bell curve compresses horizontally Its slope seems to steepen, increasing more rapidly to (0, ) and decreasing more rapidly from (0, ) Copyright 05 Pearson Education, Inc

21 Section 5 Equation Solving and Modeling 5 68 (a and b) (c) X = 80 [ 0, 50] by [0, 500] (d) y = 874# 0 x and y = ln x (e) [ 0, 50] by [0, 500] 6=874 0 x = 0x (f) ln (0 x )=ln 6 =ln 6-ln xln0=ln 6-ln 874 x = [ 0, 50] by [0, 500] Y = 65 ln 6 - ln 874 ln0 # X = 65 [ 0, 50] by [0, 500] L 0405 Y = 80 u 69 Let = 0 n, u, v 7 0 v log u = log 0 n v log u-log v=n log 0 log u-log v=n()=n For the initial expression to be true, either u and v are both powers of ten, or they are the same constant k multiplied by powers of 0 (ie, either u=0 k and v=0 m or u = a # 0 k and v = a # 0 m, 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 [0, 0] by [ 5, ] 7 Since T 0 L 6656 and T m = 45, we have ( )e -kt = 6656 * (09770) t 6656e -kt = 6656 * (09770) t 6656 e -kt = 6656 * (09770)t e -kt = * (09770) t ln e -kt =ln ( # (09770) t ) -kt=ln () + ln (09770) t -kt=0 + t ln (09770) k= - ln (09770) 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 One possible answer: We map our data so that all points (x, y) are plotted as (ln x, ln y) If these new points are linear and thus can be represented by some standard linear regression y=ax+b we make the same mapping (x : ln x, y : ln y) and find ln y=a ln x+b Using algebra and the properties of algorithms, we have: ln y=a ln x+b e ln y a ln x+b =e y= e a ln x # e = e ln # b = e b # xa e b x a = c x a, where c = e b, exactly the power regression The equations and inequalities in #7 76 must be solved graphically they cannot be solved algebraically For #77 and 78, algebraic solution is possible, although a graphical approach may be easier 74 x 066 [, 5] by [, 6] 75 x 0407 or x 09 [ 0, 0] by [ 0, 0] (b) [0, 0] by [-5, ] is a good choice The maximum energy, approximately 807, occurs when r L 79 [0, ] by [, ] Copyright 05 Pearson Education, Inc

22 ` 54 Chapter Exponential, Logistic, and Logarithmic Functions x 6 75 (approx) 77 x (approx) 78 log x - log 7 0, so log(x/9) 7 0 Then x =, 79 log(x + ) - log 6 6 0, so log x Then x =, so x + 6 6, or x 6 5 The 6 original equation also requires that x + 7 0, so the solution is - 6 x 6 5 Section 6 Mathematics of Finance Exploration k ` A A approaches a limit of about 05 y=000e is an upper bound (and asymptote) for A(x) A(x) approaches, but never equals, this bound Quick Review 6 00 # 005 = 7 50 # 005 = 75 4 so x 7 9 [, ] by [, 8] [ 40, 0] by [, 4] 0 ` ` ` ` ` ` ` ` 05 ` # 75% = 85% # 65% L 0547% 78 5 = 065 = 65% = 05 = 5% x=48 gives x= x=764 gives x= (+005)=5 dollars (+045)=550 dollars Section 6 Exercises Compound interest: A=500(+007) 6 $50; simple interest: A=500(+007(6)) $0 Compound interest: A=00(+008) 4 $4556; simple interest: A=00(+008(4)) $44 Compound interest: A=,000(+0075) 7 $9,90859; simple interest: A=,000(+0075(7)) $8,00 4 Compound interest: A=5,500(+0095) $46,05758; simple interest: A=5,500(+0095()) $,70 5 A=500a $7 4 b 6 A=500a $ b 7 A=40,500a $86,4966 b 8 A=5,00a $77,76569 b 9 A=50e (0054)(6) $78 0 A=50e (006)(8) $5507 A=,000e (007)(0) $0,404 A=8875e (0044)(5) $6,6697 a FV= 500 # b - $4, a FV= 00 # b - $0, a b - 5 FV= 450 # $70, a FV= 60 # b - $456, a PV= 857 # b 0047 $4,5 - a PV= 8568 # b 0065 $9,7690 Copyright 05 Pearson Education, Inc

23 Section 6 Mathematics of Finance 55 (8,000)a PV # 0054 i 9 R= $94 - ( + i) -n = b - a b (54,000)a PV # 007 i 0 R= $ ( + i) -n = b - a b In # 4, the time must be rounded up to the end of the next compounding period 8 Solve 00 a t (05) 4t =,so 4 b = 450: 46 ln(8/46) t= 66 years round to 6 years 4 ln 05 L 9 months (the next full compounding period) Solve 8000 a t (0075) t =, so b = 6,000: ln t= 77 years round to 7 years ln 0075 L 9 months (the next full compounding period) Solve 5,000 a t (0067) t =, so b = 45,000: ln t= 7 years round to years ln 0067 L 9 months (the next full compounding period) Note: A graphical solution provides t L 78 years round to years 0 months 4 Solve 5 a t (0) 4t =5, so 4 b = 75: ln 5 t= 57 years round to years 4 ln 0 L 9 months (the next full compounding period) 5 Solve,000a + + r, so r 0% 65 = a 7 /85 44 b 6 Solve 8500 a + r ()(5) + r b = # 8500:, so r 7% = /60 7 Solve 46 (+r) 6 =: +r= a 0 /6,so 7 b r 707% 8 Solve 8 (+r)8=5: +r= a 5 /8, so r 49% 8 b In #9 and 0, the time must be rounded up to the end of the next compounding period 4t ln 9 Solve a + b = : t= 4 4 ln 0475 L 4 round to years months 0 Solve a t b = : ln t= round to 7 years 8 ln ( /) L 76 months r (65)(5) 65 b = 6,500: For # 4, use the formula S=Pe rt Time to double: Solve =e 009t, leading to t= ln 7706 years After 5 years: 009 S=,500e (009)(5) $48,78 Time to double: Solve =e 008t, leading to t= ln 8664 years After 5 years: 008 S=,500e (008)(5) $07,9080 APR: Solve =e 4r, leading to r= ln 7% 4 After 5 years: S=9500e (07)(5) $7,866 (using the exact value of r) 4 APR: Solve =e 6r, leading to r= ln 55% 6 After 5 years: S=6,800e (055)(5) $95,055 (using the exact value of r) In #5 4, the time must be rounded up to the end of the next compounding period (except in the case of continuous compounding) 5 Solve a t, which 4 b = : t = ln 4 ln 0 L 74 rounds to 7 years 6 months 6 Solve a t, which 4 b = : t = ln 4 ln 0 L 875 rounds to 9 years (almost by 8 years 9 months) 7 Solve +007t=: t = which rounds to 007 L 49, 5 years 8 Solve 07 t ln =: t=, which rounds ln 07 L 04 to years 9 Solve a t, which 4 b = : t = ln 4 ln 075 L 999 rounds to 0 years 40 Solve a t b = : t = ln ln( + 007/) 99, which rounds to 0 years 4 Solve e 007t =: t= ln 990 years 007 For #4 45, observe that the initial balance has no effect on the APY 4 APY= a b - L 64% 4 APY= a b - L 59% 44 APY=e % 45 APY= a b - L 480% 46 The APYs are a and b - L 56% 4 b a L 5984% So, the better investment 4 is 5% compounded quarterly Copyright 05 Pearson Education, Inc

24 56 Chapter Exponential, Logistic, and Logarithmic Functions 47 The APYs are 5 and e % 8 % = 55% last payment will be less than $050 A reasonable 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, For #48 5, use the formula S= R ( + i)n - 08, and multiplying by $050, giving about $85050 To figure the exact amount of the final payment, i (0) i= = and R=50, so solve 86,000=050 +R (0) 7 00 (00605) ()(5) - S=50 L $4,46 (the present value of the first 7 payments, plus the present value of a payment of R dollars 7 months 055 from now) This gives a final payment of R $ i= = 009» and R=50, so (b) The total amount of the payments under the original (09) ()(0) - plan is 60 # $8846 = $8,45960 The total using S=50 L $80, the higher payments is 7 # $050 = $80,660 (or i= solve ; 7 # $050 + $84657 = $80,9657 if we use the correct amount of the final payment) a difference a + 04 ( * 0) of $7,85960 (or $8,060 using the correct final b - - payment) 50,000=R to obtain 57 (a) After 0 years, the remaining loan balance is 04 (0) 0-86,000(0) L $80,875 R $94 per month (round up, since $94 will not 00 be adequate) (this is the future value of the initial loan balance, 0045 minus the future value of the loan payments) With 5 i= = 00075; solve $050 payments, the time required is found by solving (0075) ()(0) - (0) -n - 80,875=050 ; this leads to 0,000=R to obtain R $ (0) n 0487, so n 456 months, or about per month (round up, since $580 will not be adequate) (additional) years The mortgage will be paid off - ( + i) -n after a total of years months, with the final payment being less than $050 A reasonable estimate of For #5 55, use the formula A=R i the final payment is (06)($050) $6000 (see the 5 i= = ; solve previous problem); to figure the exact amount, solve 9000= R - (00665)-()(4) to obtain R $95 80,875 = (0)-45 +R(0) 46, per month which gives a final payment of R $ (b) The original plan calls for a total of $8,45960 in 5 i= = ; solve payments; this plan calls for 0 # $ = R - (008547)-()() 46 # $050=$59,450 (or 0 # $8846+ to obtain R $ # $050+$669=$59,00) a savings per month (round up, since $457 will not be adequate) of $59,00640 (or $59,4947) One possible answer: The APY is the percentage increase 54 i= = ; solve from the initial balance S(0) to the end-of-year balance 86,000= R - S(); specifically, it is S()/S(0)- Multiplying the initial balance by P results in the end-of-year balance being (00797)-()(0) to obtain R $ multiplied by the same amount, so that the ratio remains per month (round up, since $67656 will not be adequate) unchanged Whether we start with a $ investment, or a i= = ; solve 00,000= $000 investment, APY= a + r k R - k b - (007708)-()(5) to obtain R $8569 per 59 One possible answer: The APR will be lower than the APY (except under annual compounding), so the bank s month (round up, since $8568 will not be adequate) offer looks more attractive when the APR is given 0 56 (a) With i=, solve = 00 Assuming monthly compounding, the APY is about 4594%; quarterly and daily compounding give approximately 4577% and 460%, respectively - (0) -n 86,000=050 ; this leads to One possible answer: Some of these situations involve (0) n = - 860, so n = 9 counting things (eg, populations), so that they can only 05 take on whole number values exponential models which months, or about 4 years The mortgage will be predict, eg, 497 fish, have to be interpreted in light of paid off after 7 months (4 years, 4 months) The this fact Copyright 05 Pearson Education, Inc