Chapter 3 Exponential, Logistic, and Logarithmic Functions

Transcription

1 ch0_p_6qxd /8/0 :5 PM Page 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 b x y = x [, ] by [, 6] [, ] by [, 6] y = a 5 b x y =5 x [, ] by [, 6] [, ] by [, 6] Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

2 ch0_p_6qxd /8/0 :5 PM Page Chapter Exponential, Logistic, and Logarithmic Functions Exploration [, ] by [, 8] [, ] by [, 8] [, ] by [, 8] [, ] by [, 8] [, ] by [, 8] fx = x fx = x gx = e 0x fx = x gx = e 05x fx = x gx = e 06x fx = x gx = e 07x fx = x gx = e 08x , since ( ) 5 = 578 0, since () =95 Section Exercises Not an exponential function because the base is variable and the exponent is constant It is a power function Exponential function, with an initial value of and base of Exponential function, with an initial value of and base of 5 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 # -/ = f(x) = # a x b g(x) = # a x b 8 a 6 b 5 f(0) = # 5 0 = # = f(x) = # () x = # x/ 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 / =( ) / = =9 5/ =( ) 5/ = 5 = 5 [, ] by [, 8] [, 7] by [, 8] 6 Translate f(x)= x by units to the left Alternatively, g(x)= x± = # x =8 # x =8 # f(x), so it can be obtained by vertically stretching f(x) by a factor of 8 [ 7, ] by [, 8] Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

3 ch0_p_6qxd /8/0 :5 PM Page 5 Section Exponential and Logistic Functions 5 7 Reflect f(x) = 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 [, ] Horizontally shrink f(x) = e x by a factor of, vertically stretch by a factor of, and shift down one unit [, 7] by [ 5, 5] 9 Vertically stretch f(x) = 05 x by a factor of and then shift 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 [, ] 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 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 0 Pearson Education, Inc Publishing as Addison-Wesley

4 ch0_p_6qxd /8/0 :5 PM Page 6 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 + = (x + ) = ( ) x + = 9 x + 0 y = y, since # x = x = + x = x y-intercept: (0, ) Horizontal asymptotes: y=0, y = [ 0, 0] by [ 5, 5] y-intercept: (0, ) Horizontal asymptotes: y = 0, y = 8 6 [, ] 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 7 [ 5, 0] by [ 5, 0] y-intercept: (0, ) Horizontal asymptotes: y = 0, y = 6 [ 5, 0] by [ 5, 0] 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 0 Pearson Education, Inc Publishing as Addison-Wesley

5 ch0_p_6qxd /8/0 :5 PM Page 7 Section Exponential and Logistic Functions 7 8 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 9 50 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 5, refer to Example 7 on page 85 in the text 5 Let P(t) be Austin s population t years after 990 Then with exponential growth, P(t) = P 0 b t where P 0 = 65,6 From Table 7, P(0) = 65,6 b 0 = 656,56 So, b = [, ] by [, 9] [, ] by [, 7] [, 7] by [, 8] 656,56 B 0 65,6 L 050 Solving graphically, we find that the curve y = 65,6(050) t intersects the line y=800,000 at t L 575 Austin s population passed 800,000 in 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(0)=6,90 b 0 =7,70 So, 7,70 b = B 0 6,90 L 08 Solving graphically, we find that the curve y = 6,90(08) t intersects the line y=800,000 at t L 00 Columbus s population will pass 800,000 in 00 5 Using the results from Exercises 5 and 5, we represent Austin s population as y=65,6(050) t and Columbus s population as y=6,90(08) t Solving graphically, we find that the curves intersect at t L 5 The two populations were equal, at 7,86, in 00 5 From the results in Exercise 5, the populations are equal at 7,86 Austin has the faster growth after that, because b is bigger (050>08) So Austin will reach million first Solving graphically, we find that the curve y=65,6(050) t intersects the line y=,000,000 at t L Austin s population will reach million in 0 55 Solving graphically, we find that the curve 79 y = intersects the line y=0 when + 0e -0009x t L 6967 Ohio s population stood at 0 million in (a) P(50) = L (50) e or,79,558 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 = 0,00, C L 567 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 6 For b>0, f(0)=b 0 = The answer is C 6 The growth factor of f(x) = a # b x is the base b The answer is A 6 With x>0, a x >b x requires a>b (regardless of whether x< or x>) The answer is B Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

6 ch0_p_6qxd /8/0 :5 PM Page 8 8 Chapter Exponential, Logistic, and Logarithmic Functions 65 (a) (b) Domain: (- q, q) Range: c e, q b Intercept: (0, 0) Decreasing on (- q, -] Increasing on [-, q) Bounded below by y = - e Local minimum at a -, - e b Asymptote y = 0 lim 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 [ 5, 5] by [, 5] f(x) = q, lim x: -q f(x) = 0 [, ] by [ 7, 5] y = 0 g(x) = -q g(x) = 0, lim x: -q 66 (a) x = ( ) =, so x = (b) x =, so x = (c) 8 x/ = x +, ( ) x/ = ( ) x + # x = x +, x = x +, x = - (d) 9 x = x +, ( ) x = x +, x = x +, x = 67 (a) y f(x) decreases less rapidly as x decreases (b) y as x increases, g(x) decreases ever more rapidly 68 c = a : To the graph of ( a ) x apply a vertical stretch by b, since f(ax + b) = ax + b = ax b = ( b )( a ) x 69 a Z 0, c = 70 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 7 Since 0<b<, lim ( + a # b x ) = q and x: -q lim ( + a # c b x ) = Thus, lim and x: - q + a # b x = 0 c lim + a # b x = c Section Exponential and Logistic Modeling Quick Review 05 % (07)() (096)(5) 5 b = 60 =, so b = ; = ; 0 6 b = 9, so b = 9 B = B 7 = 88 7 b= 0 6B b= 5B b= 06 B b= 089 7B 7 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 % 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 # 099 x (x = weeks) f(x) = 8,900 # ( - 006) x = 8,900 # 097 x (x=years) f(x) = 50,000 # ( + 007) x = 50,000 # 07 x (x=years) f(x) = 8 # ( + 005) x = 8 # 05 x (x = weeks) f(x) = 5 # ( - 006) x = 5 # 095 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) Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

7 ch0_p_6qxd /8/0 :5 PM Page 9 Section Exponential and Logistic Modeling 9 0 g 0 = -58, = 08 = r +, so The model is P(t) = 650(075) t (a) In 95: about P(5),5 In 90: about g(x)= -58 # (08) x (Decay Model) P(50),65 For #, use f(x) = f 0 # b x (b) P(t)=50,000 when t 7665 years after 890 in f Since f(5) = # b 5 0 =, so f(x) = # b x = 805, bfi= b= L 5 f(x) L # 5 5B x, The model is P(t) = 00(05) t (a) In 90: about P(0) 655 In 95: about f Since f() = # b 0 =, so f(x) = # b x = 9 P(5) (b) P(t)=0,000 when t 70 years after 90: about b = b= L 08 f(x) L # 08 B x, 980 c (a) y = 66a t/, where t is time in days For # 8, use the model f(x) = + a # b x b (b) After 8 days 0 c=0, a=, so f()= = 0, b = 0, + b (a) y = 5a t/65, where t is time in days 0 b 60b=0, b=, thus f(x)= + # a x b (b) After 78 days 5 One possible answer: Exponential and linear functions 60 are similar in that they are always increasing or always c=60, a=, so f()= =, 60 = + 96b, + b decreasing However, the two functions vary in how 60 quickly they increase or decrease While a linear function 96b=6, b=, thus f(x)= 8 + a x will increase or decrease at a steady rate over a given 8 b interval, the rate at which exponential functions increase or decrease over a given interval will vary 8 5 c=8, a=7, so f(5)=, + 7b = 6 One possible answer: Exponential functions and logistic 5 functions are similar in the sense that they are always 96 8=+bfi, bfi=96, bfi=, increasing or always decreasing They differ, however, in the sense that logistic functions have both an upper and 96 8 b=, thus f(x) L 5B L 08 lower limit to their growth (or decay), while exponential + 7 # 08 x functions generally have only a lower limit (Exponential 0 functions just keep growing) 6 c=0, a=5, so f()= = 5, 0 = b, + 5b 7 One possible answer: From the graph we see that the 5 75b =5, b = b=, 75 = doubling time for this model is years This is the time 5, B 5 L 0585 required to double from 50,000 to 00,000, from 00,000 to 00,000, or from any population size to twice that size 0 thus f(x) L Regardless of the population size, it takes years for it to + 5 # 0585 x double 0 7 c=0, a=, so f()= + b = 0, 0 = One possible answer: The number of atoms of a radioactive substance that change to a nonradioactive state in a 0b, 0b =0, b = b=,, B L 058 given time is a fixed percentage of the number of radioactive atoms initially present So the time it takes for half of 0 the atoms to change state (the half-life) does not depend thus f(x)= + # 058 x on the initial amount 9 When t=, B 00 the population doubles every hour 60 8 c=60, a=, so f(8)= + b 8 = 0, 60 = b8, 0 The half-life is about 5700 years For #, use the formula P(h)=7 # 05 h/6, where h is 90b =0, b = b=,, 8B L 087 miles above sea level 60 P(0)=7 # 05 0/6 = lb/in thus f(x)= + # 087 x P(h) = 7 # 05 h/6 intersects y = 5 when h 90 9 P(t) = 76,000(09) t miles above sea level ; P(t)=,000,000 when t 07 years, or the year 00 0 P(t) = 78,000(068) t ; P(t)=,000,000 when t years, or the year 0 [, 9] by [, 9] Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

8 ch0_p_6qxd /8/0 :5 PM Page 0 0 Chapter Exponential, Logistic, and Logarithmic Functions The exponential regression model is P(t) = 9690(0) t, where P(t) is measured in thousands of people and t is years since 900 The predicted population for Los Angeles for 007 is P(07) L 78, or,78,000 people This is an overestimate of,000 people,,000 an error of L 009 = 9%,8,000 The exponential regression model using data is P(t) = 0800(065) t, where P(t) is measured in thousands of people and t is years since 900 The predicted population for Phoenix for 007 is P(07) L, or,,000 people This is an overestimate of 679,000 people, 679,000 an error of L 0 = %,55,000 The exponential regression model using data is P(t) = 86606(076) t, where P(t) is measured in thousands of people and t is years since 900 The predicted population for Phoenix for 007 is P(07) L 597, or,957,000 people This is an overestimate of 5,000 people, 5,000 an error of L 00 = %,5,000 The equations in #5 6 can be solved either algebraically or graphically; the latter approach is generally faster 5 (a) P(0)=6 students (b) P(t)=00 when t 97 about days (c) P(t)=00 when t 690 about 7 days 6 (a) P(0)= (b) P(t)=600 when t 5 after or 5 years (c) As t : q, P(t) : 00 the population never rises above this level 7 The logistic regression model is P(x) =, where x is the e x number of years since 900 and P(x) is measured in millions of people In the year 00, x = 0, so the model predicts a population of P(0) = e (0) = e = L 06 L 0,600,000 people [ 0,0] by [0,00],0,6 8 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 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 -00x years since 800 and P is measured in millions As x : q, P(x) : 56, 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 09=+009, so the constant percentage growth rate is 009=9% The answer is C 5 The base is 08=-066, so the constant percentage decay rate is 066=66% The answer is B 55 The growth can be modeled as P(t)= # t/ Solve P(t)=000 to find t 986 The answer is D 56 Check S(0), S(), S(), 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(00) 779, or 77,900,000 people (b) The logistic model underestimates the 000 population by about 5 million, an error of around % (c) The logistic model predicted a value closer to the actual value than the exponential model, perhaps indicating a better fit 58 (a) Using the exponential growth model and the data from , Mexico's population can be represented by M(x) L 6 * 08 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 80 * 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 [0,0] by [0,500000] Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

9 ch0_p_6qxd /8/0 :5 PM Page Section Logarithmic Functions and Their Graphs eventually have a larger population Our graph indicates that this will occur at x L 8, or 8 Section Logarithmic Functions and Their Graphs Exploration [0, 00 by [ 000,0000] (b) Using logistic growth models and the same data, 658 M(x) = + 965e -00x 8090 P(x) = + 9e -006x Using this model, Mexico's population will not exceed that of the United States, confirmed by our graph [,] by [0,6] (c) According to the logistic growth models, the maximum sustainable populations are: Mexico 65 million people United States 809 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 = ex + e -x =cosh(x), so the function is even e x - e -x sinh(x) 6 (a) cosh (x) = e x + e -x e x - e -x = # e x - e -x = e x -x = tanh(x) e x + e -x + e e -x - e -(-x) (b) tanh( x)= e -x + e = e-x - e x -(-x) e -x + e x = - ex - e -x = -tanh(x), so the function is odd e x + e -x e x - e -x (c) f(x)=+tanh(x)=+ e x + e -x e x + e -x + e x - e -x e x = e x + e -x = e x + e -x e x = e x a + e -x e -x b = + e -x, which is a logistic function of c=, a=, and k= Same graph as part Quick Review 5 = = = 0 = = 5 = 6 6 = = / 8 0 / 9 0 [ 6, 6] by [, ] a / e b = e -/ a / e b = e -/ Section Exercises log = because = log 6 =0 because 6 0 = log =5 because 5 = log 8= because =8 5 log 5 5 = because 5 / = 5 6 log 6 = - because 6 -/5 = = /5 5 6 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

10 ch0_p_6qxd /8/0 :5 PM Page Chapter Exponential, Logistic, and Logarithmic Functions 7 log 0 = 8 log 0,000=log 0 = 9 log 00,000=log 0 5 =5 0 log 0 = log =log 0 / 0 = - log =log 0 / = 000 ln e = ln e = 5 ln =ln e = e 6 ln =ln e 0 =0 7 ln e=ln e / = 8 ln e = ln 7 e-7/ = -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 = 0 log 0 = e ln 6 = e loge 6 =6 e ln (/5) = e loge(/5) = /5 5 log and log and log ( ) is undefined because <0 8 log ( 5) is undefined because 5<0 9 ln and e ln 07 0 and e 0 07 ln ( 09) is undefined because 09<0 ln ( ) is undefined because <0 x=0 =00 x=0 =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) 0 f(x) is undefined for x> The answer is (c) Starting from y=ln x: translate left units Starting from y=ln x: translate up units [ 5, 5] by [, ] Starting from y=ln x: reflect across the y-axis and translate up units [, ] by [, 5] Starting from y=ln x: reflect across the y-axis and translate down units [, ] by [ 5, ] 5 Starting from y=ln x: reflect across the y-axis and translate right units [ 7, ] by [, ] 6 Starting from y=ln x: reflect across the y-axis and translate right 5 units [ 6, 6] by [, ] 7 Starting from y=log x: translate down unit [ 5, 5] by [, ] [ 5, 5] by [, ] Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

11 ch0_p_6qxd /8/0 :5 PM Page Section Logarithmic Functions and Their Graphs 8 Starting from y=log x: translate right units 5 [ 5, 5] by [, ] 9 Starting from y=log x: reflect across both axes and vertically stretch by [ 8, ] by [, ] 5 Domain: (, q) Range: ( q, q) Always increasing Not bounded Asymptote at x= lim [, 9] by [, ] f(x)=q 50 Starting from y=log x: reflect across both axes and vertically stretch by [, 8] by [, ] [ 8, 7] by [, ] 5 Starting from y=log x: reflect across the y-axis, translate right units, vertically stretch by, translate down unit 55 Domain: (, q) Range: ( q, q) Always increasing Not bounded Asymptote: x= lim f(x)=q [ 5, 5] by [, ] 5 Starting from y=log x: reflect across both axes, translate right unit, vertically stretch by, translate up unit [ 6, ] by [, ] Domain: (, q) Range: ( q, q) Always decreasing Not bounded Asymptotes: x= lim [, 8] by [, ] f(x) = -q Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

12 ch0_p_6qxd /8/0 :5 PM Page Chapter Exponential, Logistic, and Logarithmic Functions Domain: (, q) Range: ( q, q) Always decreasing Not bounded Asymptotes: x= lim [, 7] by [, ] f(x)= q # 60 I= lumens 6 The logarithmic regression model is y = ln x, where x is the year and y is the population Graph the function and use TRACE to find that x L 0 when y L 50,000,000 The population of San Antonio will reach,500,000 people in the year 0 58 [, 7] by [, ] Domain: (0, q) Range: ( q, q) Increasing on its domain No symmetry Not bounded Asymptote at x=0 lim f(x) = q [970, 00] by [600,000,,700,000] 6 The logarithmic regression model is y = ln x, where x is the year and y is the population Graph the function and use TRACE to find that x L 0 when y L 500,000 The population of Milwaukee will reach 500,000 people in the year 0 [ 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) [970, 00] by [00,000, 800,000] 6 True, by the definition of a logarithmic function 6 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 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

13 ch0_p_6qxd /8/0 :5 PM Page 5 Section Properties of Logarithmic Functions f(x) x log x Domain ( q, q) (0, q) Range (0, q) ( q, q) Intercepts (0, ) (, 0) Asymptotes y=0 x=0 [ 6, 6] by [, ] f(x) 5 x log 5 x Domain ( q, q) (0, q) Range (0, q) ( q, q) Intercepts (0, ) (, 0) Asymptotes y=0 x=0 log 09009, log (000) log 5=log a 0 =log 0-log -000 b = log 6=log = log 0 log =log 5 =5 log 5055 log 6=log 6 =6 log log 5=log 5 = log 5= log a 0 b =(log 0-log ) 979 log 0 = log ( # 0) = log + log 0 L 6006 log 50 = log a 00 b = log 00 - log L The list consists of,,, 5, 8, 6, 0, 5,, 0, 50, 6, and 80 Exploration False False; log (7x)=log 7+log x True True x 5 False; log =log x-log 6 True 7 False; log 5 x =log 5 x+log 5 x= log 5 x 8 True [ 6, 6] by [, ] e 7 b= e The point that is common to both graphs is (e, e) [ 7, 57] by [, ] 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 7 Reflect across the x-axis Section Properties of Logarithmic Functions Exploration log ( # ) 09009, log +log log a , log 8-log b Quick Review log 0 = ln e = ln e = log 0 = x 5 y - 5 x 5 y ( ) =x y x y - = 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 ) / =(x 8 ) / (y ) / = (u v - ) / - u v 9 = (7 u 6 v - 6 / ) u v - = u 0 (x - y ) - (x y - ) - = x y -6 x -9 y = x 6 y Section Exercises ln 8x=ln 8+ln x= ln +ln x ln 9y=ln 9+ln y= ln +ln y log =log -log x x log =log -log y y x 6 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

14 ch0_p_6qxd /8/0 :5 PM Page 6 6 Chapter Exponential, Logistic, and Logarithmic Functions 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 =log 000+log x =+ log x x log = (log x-log y)= log x - B y log y x ln = (ln x-ln y)= ln x - y ln y log x+log y=log xy 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 log y-log z=log y -log z=log a y z b log (xy)- log (yz)=log (x y )-log (y z ) =log a x y y z b = log a x y z b ln (x y)+ ln (yz )=ln (x 9 y )+ln (y z ) =ln (x 9 y 5 z ) In # 8, natural logarithms are shown, but common (base-0) logarithms would produce the same results ln 7 ln L 807 ln 9 ln 5 L 895 ln 75 5 L 87 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 ln(c - d) log 5 (c-d)= ln 5 log x log x= log log x log x= log log(x + y) 5 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 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 07 [, 0] by [, ] 0 Starting from g(x)=ln x: vertically shrink by a factor /ln 7 05 [, 0] by [, ] Starting from g(x)=ln x: reflect across the x-axis, then vertically shrink by a factor /ln 09 [, 0] by [, ] Starting from g(x)=ln x: reflect across the x-axis, then shrink vertically by a factor of /ln 5 L 06 [, 0] by [, ] log(x + y) = - log (b): [ 5, 5] by [, ], with Xscl= and Yscl= Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

15 ch0_p_6qxd /8/0 :5 PM Page 7 Section Properties of Logarithmic Functions 7 (graph y=ln(-x)/ ln ) (c): [, 8] by [, ], with Xscl= and Yscl= (graph y=ln(x-)/ ln 6) 5 (d): [, 8] by [, ], with Xscl= and Yscl= (graph y=ln(x-)/ ln 05) 6 (a): [ 8, ] by [ 8, 8], with Xscl= and Yscl= (graph y=ln(-x)/ ln 07) 7 [, 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 () 8 9 [, 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 log / [ 0, 0] by [, ] f(x)=q, lim x: q ln a b f(x)=q, 50 Domain: (0, q) Range: ( q, q) Always increasing Asymptote: x=0 lim [, 9] by [, 8] 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 + 5 = log (b) R=log = log I 5 log 0005(0)= 009, so = I= # L 9665 lumens I 5 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 ( # )=log +log by the product rule The answer is B 60 log 9 6=(ln 6)/(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 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

16 ch0_p_6qxd /8/0 :5 PM Page 8 8 Chapter Exponential, Logistic, and Logarithmic Functions 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) 9,66 (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 6 (a) f(x)=8095 # x 0 (b) f(9)=8095 (9) 0 60 (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) (d) log r=( 00) log (50)+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 log = log log8 = log( ) = log log9 = log( ) = log log = log + log = log + log log5 = loga 0 * b = log0 - log + log = - log + log log6 = log( ) = log log8 = log + log9 = log + log log0 = log(0 * ) = log0 + log = + log log = log( * 8) = log + log log5 log7 = log( ) = log log0 = log(0 * ) = log0 + log = + log log = log( 5 ) = 5log log6 = log + log9 = log + log log0 = log(0 * ) = log0 + log = + log log5 = log5 + log9 = - log + log log8 = log + log6 = log + log log50 log5 = log + log7 = log + log log60 = log(0 * 6) = log0 + log6 = + log + log log6 = log( 6 ) = 6log log7 = log8 + log9 = log + log log75 [, ] by [6, 8] log5 = log(0/) = log0 - log = - log log6 = log + log = loga 00 b = loga 00 b = loga 00 * b log80 = log(0 * 8) = log0 + log8 = + log log8 = log( ) = log = log00 - log = - log = log00 - log = - log = - log + log log90 = log(0 * 9) = log0 + log9 = + log log96 = log + log = log + 5log Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

17 ch0_p_6qxd /8/0 :5 PM Page 9 Section 5 Equation Solving and Modeling 9 For #67 68, 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 70 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 0 log x log e Thus, y is a constant function 7 Vertical asymptote: x= lim f(x)=q One-to one, hence invertible (f (x)= ) Section 5 Equation Solving and Modeling Exploration log ( # 0) L 6006 log ( # 0 ) L 6006 log ( # 0 ) L 6006 log ( # 0 ) L 6006 log ( # 0 5 ) L log ( # 0 6 ) L log ( # 0 7 ) L log ( # 0 8 ) L log ( # 0 9 ) L log ( # 0 0 ) L The integers increase by for every increase in a power of 0 The decimal parts are exactly equal # 0 0 is nine orders of magnitude greater than # 0 Quick Review 5 In #, 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 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 = * (-6) = 6 * 0 - e ex [, 9] by [, ] Domain: (, q) Range: ( q, q) Increasing Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

18 ch0_p_6qxd /8/0 :5 PM Page Chapter Exponential, Logistic, and Logarithmic Functions Section 5 Exercises For # 8, take a logarithm of both sides of the equation, when appropriate 6 a x/5 b = a x/ b = # 5 x/ = 50 5 x/ = 5 5 x/ = 5 x = x = # x/ = 96 x/ = x/ = 5/ x = 5 x = x/ = 0, so -x/ =, and therefore x = x/ = 5, so -x/ =, and therefore x = - 7 x = 0 = 0,000 8 x= 5 = 9 x - 5 = -, so x = = x =, so x = - x = ln 6 x = ln 098 = log L -6 e 005x =, so 005x = ln, and therefore x = e 005x =, so 005x = ln, and therefore x = a x/5 b = 9 a x/5 b = a b x 5 = x = 0 a x/ b = 6 a x/ b = a b x = x = 6 ln ln 06 = log 06 L 5 ln L ln L e -x =, so -x = ln, and therefore x = -ln L e -x = 5, so -x = ln 5, and therefore x = -ln 5 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) 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, or x = e Then x = e L 789 or x = -e L Write both sides as powers of 0, leaving 0 log x = 0, or x = 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, 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 Multiply both sides by or ( x ) - 6 # x + = 0 This is quadratic in x = 6 ; 6 - = ; Then ln( ; ) x = = log ln ( ; ) L ;5, leading to Multiply both sides by e x, leaving (e x ) + = 8e x, or (e x ) - 8e x + = 0 This is quadratic in e x, leading to e x = 8 ; 6 - = ; 5 Then x=ln( ; 5) L ;06 # x, leaving ( x ) + = 6 # x, x Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

19 ch0_p_6qxd /8/0 :5 PM Page 5 5 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 + = 0 = 00, x + = x = 00(x + ) 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+)]= ln, so (x-)(x+)=8, or x + x - 0 = 0 This factors to (x-)(x+5)=0, so x= (an actual solution) or x= 5 (extraneous, since x- and x+ must be positive) 8 log[(x-)(x+5)]= log, so (x-)(x+5)=9, or x + x - 9 = 0 Then x = ; + - ; = - 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 0 A kg hen weighs 000, or # 0 grams while a 0 g canary weighs #, 0 grams They differ by an order of magnitude of 7-55=5 They differ by an order of magnitude of 5 -=8 They differ by an order of magnitude of 8 Given b = 0 log I I 0 = 95 b = 0 log I I 0 = 65, = ; L 08 Section 5 Equation Solving and Modeling 5 This is quadratic in e x, leading to we seek the logarithm of the ratio I /I -5 ; 5 + e x = -5 ; 7 I = Of these two 0 log - 0 log I = b I 0 I - b numbers, only = is positive, so x = ln 0 alog I - log I b = I 0 I I 500 and therefore 00 = + 5e0x, so e 0x = 0 log = 0 50 = 006, I x = I log = ln 006 L I 0 The two intensities differ by orders of magnitude 00 and therefore 50 = + 95e-06x, so e -06x = 57, Given x = -06 ln 57 L 678 b = 0 log I = 70 I 0 b = 0 log I I 0 = 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 - log I b = 70-0 I 0 I 0 I 0 log = 60 I I log = 6 I The two intensities differ by 6 orders of magnitude 5 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 6 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 times greater 7 (a) Carbonated water: log [H + ]=9 log [H + ]= 9 [H + ]=0 9 6*0 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 8 (a) Stomach acid: log [H + ]=0 log [H + ]= 0 [H + ]=0 0 =*0 Blood: log [H + ]=7 log [H + ]= 7 [H + ]=0 7 98*0 8 [H + ] of stomach acid (b) = 0- L 5 * [H ] of blood 0 (c) They differ by an order of magnitude of 5 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

20 ch0_p_6qxd /8/0 :5 PM Page 5 5 Chapter Exponential, Logistic, and Logarithmic Functions The equations in #9 50 can be solved either algebraically or graphically; the latter approach is generally faster 9 Substituting known information into T(t)= T leaves T(t)=+70e kt m + (T 0 - T m )e -kt Using T()=50=+70e k, we have e k = 5, so k= - Solving T(t)=0 yields ln 5 L 0076 t 8 minutes 50 Substituting known information into T(t) = T leaves T(t)=65+85e kt m + (T 0 - T m )e -kt Using T(0)=0=65+85e 0k, we have e 0k = so k= ln 008 Solving 57, 0 57 T(t)=90 yields t 959 minutes 5 (a) (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 5 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 ) (b) [0, 0] by [0, 80] [0, 5] by [0, 00] [0, 0] by [0, 80] T(x) L 797 # 09 x (c) T(0)+0= C 5 (a) 57 Exponential the scatterplot of (x, y) is exactly exponential (The data can be modeled by y= # x ) (b) [0, 5] by [0, 90] [0, 5] by [0, 0] 58 Linear the scatterplot of (x, y) is exactly linear (y=x+) [0, 5] by [0, 90] T(x) L 7996 # 09 x (c) T(0) =7996 C 5 (a) [0, 0] by [0, 5] [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 at t gets large, so that T(t) approaches T m 6 x = x = 5 x-=5 x= The answer is B Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

21 ch0_p_6qxd /8/0 :5 PM Page 5 Section 5 Equation Solving and Modeling 5 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 a log T - log a = 8-6 T a log = a a = 0 =00 a The answer is E 6 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 + 578e b -050x closely matches the data, and would provide a natural cap to the population growth at approx 0 million people (Note: x=number of years since 900) The logistic regression model af(x) = + 6e -099x b matches the data well and provides a natural cap of million people (Note: x=number of years since 900) 67 (a) [, ] by [0, 0] As k increases, the bell curve stretches vertically Its height increases and the slope of the curve seems to steepen (b) [, ] 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, ) u 68 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 69 (a) r cannot be negative since it is a distance [ 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, 0] by [ 5, ] 70 Since T 0 L 6656 and T m = 5, we have (6656-5)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 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

22 ch0_p_6qxd /8/0 :5 PM Page 5 ` 5 Chapter Exponential, Logistic, and Logarithmic Functions 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 78, algebraic solution is possible, although a graphical approach may be easier 7 x x 007 or x x 6 75 (approx) 76 x (approx) 77 log x - log 7 0, so log(x/9) 7 0 Then x =, so x 7 9 [, 5] by [, 6] [0, ] by [, ] [, ] by [, 8] [ 0, 0] by [, ] 78 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 0 ` 06 0 ` 09 0 ` 05 0 ` ` ` ` ` 05 ` 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 # 75% = 85% # 65% L 057% 78 5 = 065 = 65% = 05 = 5% x=8 gives x= x=76 gives x= (+005)=5 dollars (+05)=550 dollars Section 6 Exercises A=500(+007) 6 $50 A=00(+008) $556 A=,000(+0075) 7 $9,90859 A=5,500(+0095) $6, A=500a $7 b 6 A=500a $57567 b 7 A=0,500a $86,966 b 8 A=5,00a $77,76569 b 9 A=50e (005)(6) $78 0 A=50e (006)(8) $5507 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

23 ch0_p_6qxd /8/0 :5 PM Page 55 Section 6 Mathematics of Finance 55 A=,000e (007)(0) $0,0 A=8875e (00)(5) $6,6697 a FV= 500 # b - $, a FV= 00 # b - $0, a b - 5 FV= 50 # $70, a FV= 60 # b - $56, a PV= 857 # b 007 $,5 - a PV= 8568 # b 0065 $9,7690 (8,000)a PV # 005 i 9 R= $9 - ( + i) -n = b - a b (5,000)a PV # 007 i 0 R= $07 - ( + i) -n = b - a b In #, the time must be rounded up to the end of the next compounding period 8 Solve 00 a t (05) t =,so b = 50: 6 ln(8/6) t= 66 years round to 6 years 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 = 5,000: 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 Solve 5 a t (0) t =5, so b = 75: ln 5 t= 57 years round to years ln 0 L 9 months (the next full compounding period) 5 Solve,000a + + r, so r 0% 65 = a 7 /85 b 6 Solve 8500 a + r ()(5) + r b = # 8500:, so r 7% = /60 7 Solve 6 (+r) 6 =: +r= a 0 /6,so 7 b r 707% 8 Solve 8 (+r)8=5: +r= a 5 /8, so r 9% 8 b In #9 0, the time must be rounded up to the end of the next compounding period t ln 9 Solve a + b = : t= ln 075 L round to years months 0 Solve a t b = : ln t= round to 7 years 8 ln ( /) L 76 months For #, 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) $8,78 Time to double: Solve =e 008t, leading to t= ln 866 years After 5 years: 008 S=,500e (008)(5) $07,9080 APR: Solve =e r, leading to r= ln 7% After 5 years: S=9500e (07)(5) $7,866 (using the exact value of r) 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 0, the time must be rounded up to the end of the next compounding period (except in the case of continuous compounding) 5 Solve a + 00 t round b = : t = ln ln 0 L 7 to 7 years 6 months 6 Solve a t round b = : t = ln ln 0 L 875 to 9 years (almost by 8 years 9 months) r (65)(5) 65 b = 6,500: Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

24 ch0_p_6qxd /8/0 :5 PM Page Chapter Exponential, Logistic, and Logarithmic Functions ln 7 Solve 07 t =: t= round to years ln 07 L 0 - ( + i) -n For #5 5, use the formula A=R i 8 Solve a t round b = : t = ln ln 075 L i= = ; solve to 0 years 9 Solve a t b = : t = 9000= R - (00665)-()() to obtain R $95 ln ln( + 007/) per month 99 round to 0 years Solve e 007t 5 i= = ; solve =: t= ln 990 years = R - (00857)-()() to obtain R $57 For #, observe that the initial balance has no effect on the APY per month (round up, since $57 will not be adequate) APY= a b L 6% 5 i= = ; solve APY= a b - L 59% 86,000= R - (00797)-()(0) to obtain R $ APY=e 006 per month (round up, since $67656 will not be adequate) - 650% APY= a b 5 i= = ; solve 00,000= - L 80% 5 The APYs are a and b R - (007708)-()(5) to obtain R $8569 per - L 56% a month (round up, since $8568 will not be adequate) b - L 598% So, the better investment 0 55 (a) With i=, solve = 00 is 5% compounded quarterly 6 The APYs are 5 - (0) -n and e % 8 % = 55% 86,000=050 ; this leads to 00 So, the better investment is 5% compounded continuously (0) n = - 860, so n 78 For #7 50, use the formula S= R ( = 9 05 i)n - months, or about years The mortgage will be i paid off after 7 months ( years, months) The 0076 last payment will be less than $050 A reasonable 7 i= = and R=50, so estimate of the final payment can be found by taking (00605) ()(5) - the fractional part of the computed value of n above, S=50 L $, , and multiplying by $050, giving about $85050 To figure the exact amount of the final payment, i= = 009» and R=50, so - (0) -7 solve 86,000=050 +R (0) 7 (09) ()(0) 00 - S=50 L $80,677 (the present value of the first 7 payments, plus the 009 present value of a payment of R dollars 7 months 0 9 i= solve ; from now) This gives a final payment of R $8657 (b) The total amount of the payments under the original a + 0 ( * 0) b plan is 60 # $886 = $8,5960 The total using - the higher payments is 7 # $050 = $80,660 (or 50,000=R to obtain 7 # $050 + $8657 = $80,9657 if we use the 0 correct amount of the final payment) a difference of $7,85960 (or $8,060 using the correct final R $9 per month (round up, since $9 will not payment) 56 (a) After 0 years, the remaining loan balance is be adequate) (0) 0-86,000(0) L $80, (this is the future value of the initial loan balance, 50 i= = 00075; solve minus the future value of the loan payments) With (0075) ()(0) - $050 payments, the time required is found by solving 0,000=R to obtain R $ (0) -n 80,875=050 ; this leads to per month (round up, since $580 will not be adequate) 00 (0) n 087, so n 56 months, or about Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley

25 ch0_p_6qxd /8/0 :5 PM Page 57 Section 6 Mathematics of Finance 57 (additional) years The mortgage will be paid off after a total of years months, with the final payment being less than $050 A reasonable estimate of the final payment is (06)($050) $6000 (see the previous problem); to figure the exact amount, solve 80,875 = (0)-5 +R(0) 6, 00 which gives a final payment of R $669 (b) The original plan calls for a total of $8,5960 in payments; this plan calls for 0 # $ # $050=$59,50 (or 0 # $ # $050+$669=$59,00) a savings of $59,0060 (or $59,97) 57 One possible answer: The APY is the percentage increase from the initial balance S(0) to the end-of-year balance S(); specifically, it is S()/S(0)- Multiplying the initial balance by P results in the end-of-year balance being multiplied by the same amount, so that the ratio remains unchanged Whether we start with a $ investment, or a $000 investment, APY= a + r k k b - 58 One possible answer: The APR will be lower than the APY (except under annual compounding), so the bank s offer looks more attractive when the APR is given Assuming monthly compounding, the APY is about 59%; quarterly and daily compounding give approximately 577% and 60%, respectively 59 One possible answer: Some of these situations involve counting things (eg, populations), so that they can only take on whole number values exponential models which predict, eg, 97 fish, have to be interpreted in light of this fact Technically, bacterial growth, radioactive decay, and compounding interest also are counting problems for example, we cannot have fractional bacteria, or fractional atoms of radioactive material, or fractions of pennies However, because these are generally very large numbers, it is easier to ignore the fractional parts (This might also apply when one is talking about, eg, the population of the whole world) Another distinction: while we often use an exponential model for all these situations, it generally fits better (over long periods of time) for radioactive decay than for most of the others Rates of growth in populations (esp human populations) tend to fluctuate more than exponential models suggest Of course, an exponential model also fits well in compound interest situations where the interest rate is held constant, but there are many cases where interest rates change over time 60 (a) Steve s balance will always remain $000, since interest is not added to it Every year he receives 6% of that $000 in interest: 6% in the first year, then another 6% in the second year (for a total of # 6%=%), then another 6% (totaling # 6%=8%), etc After t years, he has earned 6t% of the $000 investment, meaning that altogether he has # 006t=000(+006t) (b) The table is shown below; the second column gives values of 000(06) t The effects of annual compounding show up beginning in Year Not Years Compounded Compounded False The limit, with continuous compounding, is A = Pe rt = 00 e 005 L $05 6 True The calculation of interest paid involves compounding, and the compounding effect is greater for longer repayment periods 6 A = P( + r/k) kt = 50( + 007/) (6) L $00 The answer is B 6 Let x=apy Then +x=(+006/) 067 So x 0067 The answer is C 65 FV=R((+i) n -)/i= 00((+00075) 0 -)/00075 $6,7 The answer is E 66 R=PV i/(-(+i) n ) =0,000(0075/)/(-(+0075/) 80 ) $095 The answer is A 67 The last payment will be $68 68 One possible answer: The answer is (c) This graph shows the loan balance decreasing at a fairly steady rate over time By contrast, the early payments on a 0-year mortgage go mostly toward interest, while the late payments go mostly toward paying down the debt So the graph of loan balance versus time for a 0-year mortgage at double the interest rate would start off nearly horizontal and more steeply decrease over time ( + i) n - 69 (a) Matching up with the formula S=R, i where i=r/k, with r being the rate and k being the number of payments per year, we find r=8% (b) k= payments per year (c) Each payment is R=$00 - ( + i) -n 70 (a) Matching up with the formula A=R, i where i=r/k, with r being the rate and k being number of payments per year, we find r=8% (b) k= payments per year (c) Each payment is R=$00 Copyright 0 Pearson Education, Inc Publishing as Addison-Wesley