Exponential, Logistic, and Logarithmic Functions

CHAPTER 3 Eponential, Logistic, and Logarithmic Functions 3.1 Eponential and Logistic Functions 3.2 Eponential and Logistic Modeling 3.3 Logarithmic Functions and Their Graphs 3.4 Properties of Logarithmic Functions 3.5 Equation Solving and Modeling 3.6 Mathematics of Finance The loudness of a sound we hear is based on the intensit of the associated sound wave. This sound intensit is the energ per unit time of the wave over a given area, measured in watts per square meter 1W/m 2 2. The intensit is greatest near the source and decreases as ou move awa, whether the sound is rustling leaves or rock music. Because of the wide range of audible sound intensities, the are generall converted into decibels, which are based on logarithms. See page 280. 251

252 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions Chapter 3 Overview In this chapter, we stud three interrelated families of functions: eponential, logistic, and logarithmic functions. Polnomial functions, rational functions, and power functions with rational eponents are algebraic functions functions obtained b adding, subtracting, multipling, and dividing constants and an independent variable, and raising epressions to integer powers and etracting roots. In this chapter and the net one, we eplore transcendental functions, which go beond, or transcend, these algebraic operations. Just like their algebraic cousins, eponential, logistic, and logarithmic functions have wide application. Eponential functions model growth and deca over time, such as unrestricted population growth and the deca of radioactive substances. Logistic functions model restricted population growth, certain chemical reactions, and the spread of rumors and diseases. Logarithmic functions are the basis of the Richter scale of earthquake intensit, the ph acidit scale, and the decibel measurement of sound. The chapter closes with a stud of the mathematics of finance, an application of eponential and logarithmic functions often used when making investments. What ou ll learn about Eponential Functions and Their Graphs The Natural Base e Logistic Functions and Their Graphs Population Models... and wh Eponential and logistic functions model man growth patterns, including the growth of human and animal populations. 20 15 10 5 4 3 2 1 1 2 3 4 FIGURE 3.1 Sketch of g12 = 2. 3.1 Eponential and Logistic Functions Eponential Functions and Their Graphs The functions ƒ12 = 2 and g12 = 2 each involve a base raised to a power, but the roles are reversed: For ƒ12 = 2, the base is the variable, and the eponent is the constant 2; ƒ is a familiar monomial and power function. For g12 = 2, the base is the constant 2, and the eponent is the variable ; g is an eponential function. See Figure 3.1. DEFINITION Eponential Functions Let a and b be real number constants. An eponential function in is a function that can be written in the form ƒ12 = a # b, where a is nonzero, b is positive, and b Z 1. The constant a is the initial value of ƒ (the value at = 0), and b is the base. Eponential functions are defined and continuous for all real numbers. It is important to recognize whether a function is an eponential function. EXAMPLE 1 Identifing Eponential Functions ƒ12 = 3 is an eponential function, with an initial value of 1 and base of 3. g12 = 6-4 is not an eponential function because the base is a variable and the eponent is a constant; g is a power function. (c) h12 = -2 # 1.5 is an eponential function, with an initial value of -2 and base of 1.5.

SECTION 3.1 Eponential and Logistic Functions 253 (d) k12 = 7 # 2 - is an eponential function, with an initial value of 7 and base of 1/2 because 2 - = 12-1 2 = 11/22. (e) q12 = 5 # 6 p is not an eponential function because the eponent p is a constant; q is a constant function. Now tr Eercise 1. One wa to evaluate an eponential function, when the inputs are rational numbers, is to use the properties of eponents. EXAMPLE 2 Computing Eponential Function Values for Rational Number Inputs For ƒ12 = 2, (c) ƒ142 = 2 4 = 2 # 2 # 2 # 2 = 16 ƒ102 = 2 0 = 1 ƒ1-32 = 2-3 = 1 2 3 = 1 8 = 0.125 (d) (e) ƒa 1 2 b = 21/2 = 12 = 1.4142 Á ƒa - 3 2 b = 2-3/2 = 1 2 3/2 = 1 22 = 1 3 18 = 0.35355 Á Now tr Eercise 7. There is no wa to use properties of eponents to epress an eponential function s value for irrational inputs. For eample, if ƒ12 = 2, ƒ1p2 = 2 p, but what does 2 p mean? Using properties of eponents, 2 3 = 2 # 2 # 2, 2 3.1 = 2 31/10 = 22 10 31. So we can find meaning for 2 p b using successivel closer rational approimations to p as shown in Table 3.1. Table 3.1 Values of ƒ12 = 2 for Rational Numbers Approaching P 3.14159265... 3 3.1 3.14 3.141 3.1415 3.14159 2 8 8.5... 8.81... 8.821... 8.8244... 8.82496... We can conclude that ƒ1p2 = 2 p L 8.82, which could be found directl using a grapher. The methods of calculus permit a more rigorous definition of eponential functions than we give here, a definition that allows for both rational and irrational inputs. The wa eponential functions change makes them useful in applications. This pattern of change can best be observed in tabular form.

254 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions ( 2, 4/9) ( 1, 4/3) ( 2, 128) [ 2.5, 2.5] b [ 10, 50] ( 1, 32) (0, 4) (0, 8) [ 2.5, 2.5] b [ 25, 150] (2, 36) (1, 12) (1, 2) (2, 1/2) FIGURE 3.2 Graphs of g12 = 4 # 3 and h12 = 8 # (1/42. (Eample 3) EXAMPLE 3 Finding an Eponential Function from Its Table of Values Determine formulas for the eponential functions g and h whose values are given in Table 3.2. Table 3.2 Values for Two Eponential Functions -2-1 0 1 g12 4/9 4/3 4 12 * 3 * 3 * 3 * 3 h12 128 32 2 36 1/2 SOLUTION Because g is eponential, g12 = a # b. Because g102 = 4, the initial value a is 4. Because g112 = 4 # b 1 = 12, the base b is 3. So, Because h is eponential, h12 = a # b. Because h102 = 8, the initial value a is 8. Because h112 = 8 # b 1 = 2, the base b is 1/4. So, 8 2 g12 = 4 # 3. * 1 4 * 1 4 * 1 4 * 1 4 h12 = 8 # a 1 4 b. Figure 3.2 shows the graphs of these functions pass through the points whose coordinates are given in Table 3.2. Now tr Eercise 11. Table 3.3 Values for a General Eponential Function ƒ12 a # b a # b -2 ab -2 * b -1 ab -1 * b 0 a * b 1 ab * b 2 ab 2 Observe the patterns in the g12 and h12 columns of Table 3.2. The g12 values increase b a factor of 3 and the h12 values decrease b a factor of 1/4, as we add 1 to moving from one row of the table to the net. In each case, the change factor is the base of the eponential function. This pattern generalizes to all eponential functions as illustrated in Table 3.3. In Table 3.3, as increases b 1, the function value is multiplied b the base b. This relationship leads to the following recursive formula. Eponential Growth and Deca For an eponential function ƒ12 = a # b and an real number, ƒ1 + 1) = b # ƒ12. If a 7 0 and b 7 1, the function f is increasing and is an eponential growth function. The base b is its growth factor. If a 7 0 and b 6 1, f is decreasing and is an eponential deca function. The base b is its deca factor.

SECTION 3.1 Eponential and Logistic Functions 255 In Eample 3, g is an eponential growth function, and h is an eponential deca function. As increases b 1, grows b a factor of 3, and h12 = 8 # (1/42 g12 = 4 # 3 decas b a factor of 1/4. Whenever the initial value is positive, the base of an eponential function, like the slope of a linear function, tells us whether the function is increasing or decreasing and b how much. So far, we have focused most of our attention on the algebraic and numerical aspects of eponential functions. We now turn our attention to the graphs of these functions. Graphs of Eponential Functions EXPLORATION 1 1. Graph each function in the viewing window 3-2, 24 b 3-1, 64. (c) (d) 4 = 5 3 = 4 2 = 3 1 = 2 Which point is common to all four graphs? Analze the functions for domain, range, continuit, increasing or decreasing behavior, smmetr, boundedness, etrema, asmptotes, and end behavior. 2. Graph each function in the viewing window 3-2, 24 b 3-1, 64. 2 = a 1 1 = a 1 2 b 3 b (c) 3 = a 1 4 b (d) 4 = a 1 5 b Which point is common to all four graphs? Analze the functions for domain, range, continuit, increasing or decreasing behavior, smmetr, boundedness, etrema, asmptotes, and end behavior. We summarize what we have learned about eponential functions with an initial value of 1. f () = b b > 1 (1, b) f () = b 0 < b < 1 (0, 1) (0, 1) (1, b) FIGURE 3.3 Graphs of ƒ12 = b for b 7 1 and 0 6 b 6 1. Eponential Functions ƒ12 b Domain: All reals Range: 10, q2 Continuous No smmetr: neither even nor odd Bounded below, but not above No local etrema Horizontal asmptote: = 0 No vertical asmptotes If b 7 1 (see Figure 3.3a), then ƒ is an increasing function, lim ƒ12 = 0 and lim ƒ12 = q. : -q : q If 0 6 b 6 1 (see Figure 3.3b), then ƒ is a decreasing function, lim ƒ12 = q and lim ƒ12 = 0. : -q : q

256 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions The translations, reflections, stretches, and shrinks studied in Section 1.5, together with our knowledge of the graphs of basic eponential functions, allow us to predict the graphs of the functions in Eample 4. EXAMPLE 4 Transforming Eponential Functions Describe how to transform the graph of ƒ12 = 2 into the graph of the given function. Sketch the graphs b hand and support our answer with a grapher. (c) k12 = 3 # 2 g12 = 2-1 h12 = 2 - SOLUTION The graph of g12 = 2-1 is obtained b translating the graph of ƒ12 = 2 b 1 unit to the right (Figure 3.4a). We can obtain the graph of h12 = 2 - b reflecting the graph of ƒ12 = 2 across the -ais (Figure 3.4b). Because 2 - = 12-1 2 = 11/22, we can also think of h as an eponential function with an initial value of 1 and a base of 1/2. (c) We can obtain the graph of k12 = 3 # 2 b verticall stretching the graph of ƒ12 = 2 b a factor of 3 (Figure 3.4c). Now tr Eercise 15. [ 4, 4] b [ 2, 8] [ 4, 4] b [ 2, 8] [ 4, 4] b [ 2, 8] (c) FIGURE 3.4 The graph of shown with,, and (c) k12 = 3 # 2 ƒ12 = 2 g12 = 2-1 h12 = 2 -. (Eample 4) The Natural Base e The function ƒ12 = e is one of the basic functions introduced in Section 1.3, and is an eponential growth function. BASIC FUNCTION The Eponential Function [ 4, 4] b [ 1, 5] FIGURE 3.5 The graph of ƒ12 = e. ƒ12 = e Domain: All reals Range: 10, q2 Continuous Increasing for all No smmetr Bounded below, but not above No local etrema Horizontal asmptote: = 0 No vertical asmptotes End behavior: lim and lim : q e = q : -q e = 0 Because ƒ12 = e is increasing, it is an eponential growth function, so e 7 1. But what is e, and what makes this eponential function the eponential function?

SECTION 3.1 Eponential and Logistic Functions 257 The letter e is the initial of the last name of Leonhard Euler (1707 1783), who introduced the notation. Because ƒ12 = e has special calculus properties that simplif man calculations, e is the natural base of eponential functions for calculus purposes, and ƒ12 = e is considered the natural eponential function. DEFINITION The Natural Base e e = lim a1 + 1 : q b We cannot compute the irrational number e directl, but using this definition we can obtain successivel closer approimations to e, as shown in Table 3.4. Continuing the process in Table 3.4 with a sufficientl accurate computer can show that e L 2.718281828459. f () = e k k > 0 (0, 1) (1, e k ) Table 3.4 Approimations Approaching the Natural Base e 1 10 100 1000 10,000 100,000 11 + 1/2 2 2.5... 2.70... 2.716... 2.7181... 2.71826... We are usuall more interested in the eponential function ƒ12 = e and variations of this function than in the irrational number e. In fact, an eponential function can be epressed in terms of the natural base e. f () = e k k < 0 THEOREM Eponential Functions and the Base e An eponential function ƒ12 = a # b can be rewritten as ƒ12 = a # e k, (0, 1) (1, e k ) for an appropriatel chosen real number constant k. If and k 7 0, ƒ12 = a # e k a 7 0 is an eponential growth function. (See Figure 3.6a.) If and k 6 0, ƒ12 = a # e k a 7 0 is an eponential deca function. (See Figure 3.6b.) FIGURE 3.6 Graphs of ƒ12 = e k for k 7 0 and k 6 0. In Section 3.3 we will develop some mathematics so that, for an positive number b Z 1, we can easil find the value of k such that e k = b. In the meantime, we can use graphical and numerical methods to approimate k, as ou will discover in Eploration 2. EXPLORATION 2 Choosing k so that e k 2 1. Graph ƒ12 = 2 in the viewing window 3-4, 44 b 3-2, 84. 2. One at a time, overla the graphs of g12 = e k for k = 0.4, 0.5, 0.6, 0.7, and 0.8. For which of these values of k does the graph of g most closel match the graph of ƒ? 3. Using tables, find the 3-decimal-place value of k for which the values of g most closel approimate the values of ƒ.

258 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions EXAMPLE 5 Transforming Eponential Functions Describe how to transform the graph of ƒ12 = e into the graph of the given function. Sketch the graphs b hand and support our answer with a grapher. g12 = e 2 h12 = e - (c) k12 = 3e [ 4, 4] b [ 2, 8] SOLUTION The graph of g12 = e 2 is obtained b horizontall shrinking the graph of ƒ12 = e b a factor of 2 (Figure 3.7a). We can obtain the graph of h12 = e - b reflecting the graph of ƒ12 = e across the -ais (Figure 3.7b). (c) We can obtain the graph of k12 = 3e b verticall stretching the graph of ƒ12 = e b a factor of 3 (Figure 3.7c). Now tr Eercise 21. [ 4, 4] b [ 2, 8] [ 4, 4] b [ 2, 8] (c) FIGURE 3.7 The graph of ƒ12 = e shown with g12 = e 2, h12 = e -, and (c) k12 = 3e. (Eample 5) Logistic Functions and Their Graphs Eponential growth is unrestricted. An eponential growth function increases at an ever-increasing rate and is not bounded above. In man growth situations, however, there is a limit to the possible growth. A plant can onl grow so tall. The number of goldfish in an aquarium is limited b the size of the aquarium. In such situations the growth often begins in an eponential manner, but the growth eventuall slows and the graph levels out. The associated growth function is bounded both below and above b horizontal asmptotes. DEFINITION Logistic Growth Functions Let a, b, c, and k be positive constants, with b 6 1. A logistic growth function in is a function that can be written in the form ƒ12 = c 1 + a # b or ƒ12 = where the constant c is the limit to growth. c 1 + a # e -k Aliases for Logistic Growth Logistic growth is also known as restricted, inhibited, or constrained eponential growth. If b 7 1 or k 6 0, these formulas ield logistic deca functions. Unless otherwise stated, all logistic functions in this book will be logistic growth functions. B setting a = c = k = 1, we obtain the logistic function ƒ12 = 1 1 + e -. This function, though related to the eponential function e, cannot be obtained from b translations, reflections, and horizontal and vertical stretches and shrinks. So we give the logistic function a formal introduction: e

SECTION 3.1 Eponential and Logistic Functions 259 BASIC FUNCTION The Logistic Function [ 4.7, 4.7] b [ 0.5, 1.5] FIGURE 3.8 The graph of ƒ12 = 1/11 + e - 2. 1 ƒ12 = 1 + e - Domain: All reals Range: 10, 12 Continuous Increasing for all Smmetric about 10, 1/22, but neither even nor odd Bounded below and above No local etrema Horizontal asmptotes: = 0 and = 1 No vertical asmptotes End behavior: lim ƒ12 = 0 and lim ƒ12 = 1 : -q : q All logistic growth functions have graphs much like the basic logistic function. Their end behavior is alwas described b the equations lim ƒ12 = 0 : -q and lim ƒ12 = c, : q where c is the limit to growth (see Eercise 73). All logistic functions are bounded b their horizontal asmptotes, = 0 and = c, and have a range of 10, c2. Although ever logistic function is smmetric about the point of its graph with - coordinate c/2, this point of smmetr is usuall not the -intercept, as we can see in Eample 6. EXAMPLE 6 Graphing Logistic Growth Functions Graph the function. Find the -intercept and the horizontal asmptotes. ƒ12 = 8 1 + 3 # 0.7 SOLUTION The graph of ƒ12 = 8/11 + 3 # 0.7 2 is shown in Figure 3.9a. The -intercept is Because the limit to growth is 8, the horizontal asmptotes are = 0 and = 8. The graph of g12 = 20/11 + 2e -3 2 is shown in Figure 3.9b. The -intercept is g102 = ƒ102 = g12 = 8 1 + 3 # 0.7 0 = 8 1 + 3 = 2. 20 1 + 2e -3 # 0 = 20 1 + 2e -3 20 1 + 2 = 20/3 L 6.67. Because the limit to growth is 20, the horizontal asmptotes are = 0 and = 20. Now tr Eercise 41.

260 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions [ 10, 20] b [ 2, 10] [ 2, 4] b [ 5, 25] FIGURE 3.9 The graphs of ƒ12 = 8/(1 + 3 # 0.7 2 and g12 = 20/(1 + 2e -3 2. (Eample 6) Population Models Eponential and logistic functions have man applications. One area where both tpes of functions are used is in modeling population. Between 1990 and 2000, both Phoeni and San Antonio passed the 1 million mark. With its Silicon Valle industries, San Jose, California, appears to be the net U.S. cit destined to surpass 1 million residents. When a cit s population is growing rapidl, as in the case of San Jose, eponential growth is a reasonable model. A Note on Population Data When the U.S. Census Bureau reports a population estimate for a given ear, it generall represents the population at the middle of the ear, or Jul 1. We will assume this to be the case when interpreting our results to population problems unless otherwise noted. EXAMPLE 7 Modeling San Jose s Population Using the data in Table 3.5 and assuming the growth is eponential, when will the population of San Jose, California, surpass 1 million persons? SOLUTION Model Let P1t2 be the population of San Jose t ears after Jul 1, 2000. (See margin note.) Because P is eponential, P1t2 = P 0 # b t, where P 0 is the initial (2000) population of 898,759. From Table 3.5 we see that P172 = 898759b 7 = 939899. So, b = B 7 939,899 898,759 L 1.0064 and P1t2 = 898,759 # 1.0064 t. Solve Graphicall Figure 3.10 shows that this population model intersects = 1,000,000 when the independent variable is about 16.73. Interpret Because 16.73 r after mid-2000 is in the first half of 2017, according to this model the population of San Jose will surpass the 1 million mark in earl 2017. Now tr Eercise 51. Table 3.5 The Population of San Jose, California Year Population 2000 898,759 2007 939,899 Source: U.S. Census Bureau. Intersection X=16.731494 Y=1000000 [ 10, 30] b [800 000, 1 100 000] FIGURE 3.10 A population model for San Jose, California. (Eample 7)

SECTION 3.1 Eponential and Logistic Functions 261 While San Jose s population is soaring, in other major cities, such as Dallas, the population growth is slowing. The once sprawling Dallas is now constrained b its neighboring cities. A logistic function is often an appropriate model for restricted growth, such as the growth that Dallas is eperiencing. Intersection X=84.513263 Y=1000000 EXAMPLE 8 Modeling Dallas s Population Based on recent census data, a logistic model for the population of Dallas, t ears after 1900, is as follows: P1t2 = 1,301,642 1 + 21.602e -0.05054t According to this model, when was the population 1 million? SOLUTION Figure 3.11 shows that the population model intersects = 1,000,000 when the independent variable is about 84.51. Because 84.51 r after mid-1900 is at the beginning of 1985, if Dallas s population has followed this logistic model, its population was 1 million then. Now tr Eercise 55. [0, 120] b [ 500 000, 1 500 000] FIGURE 3.11 A population model for Dallas, Teas. (Eample 8) QUICK REVIEW 3.1 (For help, go to Sections A.1 and P.1.) Eercise numbers with a gra background indicate problems that the authors have designed to be solved without a calculator. In Eercises 1 4, evaluate the epression without using a calculator. 1. 23-216 2. 3 125 A 8 3. 27 2/3 4. 4 5/2 In Eercises 5 8, rewrite the epression using a single positive eponent. 5. 12-3 2 4 6. 13 4 2-2 7. 1a -2 2 3 8. 1b -3 2-5 In Eercises 9 10, use a calculator to evaluate the epression. 9. 25-5.37824 10. 24 92.3521 SECTION 3.1 EXERCISES In Eercises 1 6, which of the following are eponential functions? For those that are eponential functions, state the initial value and the base. For those that are not, eplain wh not. 1. = 8 2. = 3 3. = 5 4. = 4 2 5. = 2 6. = 1.3 In Eercises 7 10, compute the eact value of the function for the given -value without using a calculator. 7. ƒ12 = 3 # 5 for = 0 8. ƒ12 = 6 # 3 for = -2 9. ƒ12 = -2 # 3 for = 1/3 10. ƒ12 = 8 # 4 for = -3/2 In Eercises 11 and 12, determine a formula for the eponential function whose values are given in Table 3.6. 11. ƒ12 12. g12 Table 3.6 Values for Two Eponential Functions ƒ12 g12-2 6 108-1 3 36 0 3/2 12 1 3/4 4 2 3/8 4/3

262 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions In Eercises 13 and 14, determine a formula for the eponential function whose graph is shown in the figure. 13. ƒ12 14. g12 = f() (0, 3) (2, 6) = g() (0, 2) b a1, 2 e In Eercises 15 24, describe how to transform the graph of ƒ into the graph of g. Sketch the graphs b hand and support our answer with a grapher. -3 15. ƒ12 = 2, g12 = 2 +4 16. ƒ12 = 3, g12 = 3-17. ƒ12 = 4, g12 = 4 5-18. ƒ12 = 2, g12 = 2 19. ƒ12 = 0.5, g12 = 3 # 0.5 + 4 20., g12 = 2 # 0.6 3 ƒ12 = 0.6-2 21. ƒ12 = e, g12 = e -3 22. ƒ12 = e, g12 = -e 3-3 23. ƒ12 = e, g12 = 2e 24. ƒ12 = e, g12 = 3e 2-1 In Eercises 25 30, match the given function with its graph. Writing to Learn Eplain how to make the choice without using a grapher. 25. = 3 26. = 2-27. = -2 28. = -0.5 29. = 3 - - 2 30. = 1.5-2 In Eercises 31 34, state whether the function is an eponential growth function or eponential deca function, and describe its end behavior using limits. 31. ƒ12 = 3-2 32. ƒ12 = a 1 e b 33. ƒ12 = 0.5 34. ƒ12 = 0.75 - In Eercises 35 38, solve the inequalit graphicall. 35. 9 6 4 36. 6-7 8-37. a 1 4 b 7 a 1 3 b 38. a 1 3 b 6 a 1 2 b Group Activit In Eercises 39 and 40, use the properties of eponents to prove that two of the given three eponential functions are identical. Support graphicall. 39. 1 = 3 2+4 2 = 3 2 + 4 (c) 3 = 9 +2 40. 1 = 4 3-2 2 = 212 3-2 2 (c) 3 = 2 3-1 In Eercises 41 44, use a grapher to graph the function. Find the -intercept and the horizontal asmptotes. 12 18 41. ƒ12 = 42. ƒ12 = 1 + 5 # 0.2 1 + 2 # 0.8 16 9 43. ƒ12 = 44. g12 = 1 + 3e -2 1 + 2e - In Eercises 45 50, graph the function and analze it for domain, range, continuit, increasing or decreasing behavior, smmetr, boundedness, etrema, asmptotes, and end behavior. 45. 46. ƒ12 = 4 # 0.5 ƒ12 = 3 # 2 47. 48. ƒ12 = 5 # e - ƒ12 = 4 # e 3 5 6 49. ƒ12 = 50. ƒ12 = 1 + 2 # e - 1 + 4 # e -2 51. Population Growth Using the midear data in Table 3.7 and assuming the growth is eponential, when did the population of Austin surpass 800,000 persons? (c) (e) (d) (f) Table 3.7 Populations of Two Major U.S. Cities Cit 1990 Population 2000 Population Austin, Teas 465,622 656,562 Columbus, Ohio 632,910 711,265 Source: World Almanac and Book of Facts 2005.

SECTION 3.1 Eponential and Logistic Functions 263 52. Population Growth Using the data in Table 3.7 and assuming the growth is eponential, when would the population of Columbus surpass 800,000 persons? 53. Population Growth Using the data in Table 3.7 and assuming the growth is eponential, when were the populations of Austin and Columbus equal? 54. Population Growth Using the data in Table 3.7 and assuming the growth is eponential, which cit Austin or Columbus would reach a population of 1 million first, and in what ear? 55. Population Growth Using 20th-centur U.S. census data, the population of Ohio can be modeled b 12.79 P1t2 = 1 + 2.402e -0.0309, where P is the population in millions and t is the number of ears since April 1, 1900. Based on this model, when was the population of Ohio 10 million? 56. Population Growth Using 20th-centur U.S. census data, the population of New York state can be modeled b 19.875 P1t2 = 1 + 57.993e -0.035005t, where P is the population in millions and t is the number of ears since 1800. Based on this model, What was the population of New York in 1850? What will New York state s population be in 2015? (c) What is New York s maimum sustainable population (limit to growth)? 57. Bacteria Growth The number B of bacteria in a petri dish culture after t hours is given b B = 100e 0.693t. What was the initial number of bacteria present? How man bacteria are present after 6 hours? 58. Carbon Dating The amount C in grams of carbon-14 present in a certain substance after t ears is given b C = 20e -0.0001216t. What was the initial amount of carbon-14 present? How much is left after 10,400 ears? When will the amount left be 10 g? Standardized Test Questions 59. True or False Ever eponential function is strictl increasing. Justif our answer. 60. True or False Ever logistic growth function has two horizontal asmptotes. Justif our answer. In Eercises 61 64, solve the problem without using a calculator. 61. Multiple Choice Which of the following functions is eponential? 3 (A) ƒ12 = a 2 (B) ƒ12 = (C) ƒ12 = 2/3 (D) ƒ12 = 23 (E) ƒ12 = 8 62. Multiple Choice What point do all functions of the form ƒ12 = b 1b 7 02 have in common? (A) 11, 12 (B) 11, 02 (C) 10, 12 (D) 10, 02 (E) 1-1, -12 63. Multiple Choice The growth factor for ƒ12 = 4 # 3 is (A) 3. (B) 4. (C) 12. (D) 64. (E) 81. 64. Multiple Choice For 7 0, which of the following is true? (A) 3 7 4 (B) 7 7 5 (C) 11/62 7 (1/22 (D) 9-7 8 - (E) 0.17 7 0.32 Eplorations 65. Graph each function and analze it for domain, range, increasing or decreasing behavior, boundedness, etrema, asmptotes, and end behavior. ƒ12 = # e g12 = e- 66. Use the properties of eponents to solve each equation. Support graphicall. 2 = 4 2 3 = 27 (c) 8 /2 = 4 +1 (d) 9 = 3 +1 Etending the Ideas 67. Writing to Learn Table 3.8 gives function values for = ƒ12 and = g12. Also, three different graphs are shown. Table 3.8 Data for Two Functions ƒ12 g12 1.0 5.50 7.40 1.5 5.35 6.97 2.0 5.25 6.44 2.5 5.17 5.76 3.0 5.13 4.90 3.5 5.09 3.82 4.0 5.06 2.44 4.5 5.05 0.71 2 3 1 Which curve of those shown in the graph most closel resembles the graph of = ƒ12? Eplain our choice. Which curve most closel resembles the graph of = g12? Eplain our choice.

264 CHAPTER 3 Eponential, Logistic, and Logarithmic Functions 68. Writing to Learn Let ƒ12 = 2. Eplain wh the graph of ƒ1a + b2 can be obtained b appling one transformation to the graph of = c for an appropriate value of c. What is c? Eercises 69 72 refer to the epression ƒ1a, b, c2 = a # b c. For eample, if,, and, the epression is ƒ12, 3, 2 = 2 # 3 a = 2 b = 3 c =, an eponential function. 69. If b =, state conditions on a and c under which the epression ƒ1a, b, c2 is a quadratic power function. 70. If b =, state conditions on a and c under which the epression ƒ(a, b, c) is a decreasing linear function. 71. If c =, state conditions on a and b under which the epression ƒ(a, b, c) is an increasing eponential function. 72. If c =, state conditions on a and b under which the epression ƒ(a, b, c) is a decreasing eponential function. 73. Prove that lim c c and lim, : q 1 + a # b = c : -q 1 + a # b = 0 for constants a, b, and c, with a 7 0, 0 6 b 6 1, and c 7 0.