2. Copyright © 2004 Pearson Education, Inc. Chapter 4 Exponential and Logarithmic Functions

3. Copyright © 2004 Pearson Education, Inc. 4.1 Inverse Functions

4. Slide 4-4 Copyright © 2004 Pearson Education, Inc. Functions One-to-One Function In a one-to-one function, each x-value corresponds to only one y-value, and each y-value corresponds to only one x-value. 456456 123123 One-to-one function. 246246 7 9 11 Not a one-to-one function.

5. Slide 4-5 Copyright © 2004 Pearson Education, Inc. Example Determine whether a function defined by is one-to-one. Solution: If any y-value in a function corresponds to more than one x-value, then the function is not one-to-one. Here, a given y-value may correspond to more than one x-value. For example, both x = 2 and x =  2 corresponds to y = 4. Thus, by definition, f is not a one-to-one function.

6. Slide 4-6 Copyright © 2004 Pearson Education, Inc. Horizontal Line Test If any line intersects the graph of a function in no more than one point, then the function is one-to- one.

7. Slide 4-7 Copyright © 2004 Pearson Education, Inc. Example Solution: Since more than one different values of x leads to the same value of y, the function is not one-to-one. Solution: Since every horizontal line will intersect the graph at exactly one point, this function is one-to- one. Determine whether the graphs are one-to-one functions.

8. Slide 4-8 Copyright © 2004 Pearson Education, Inc. One-to-One Functions Test for One-to-One Functions In a one-to-one function every y-value corresponds to no more than one x-value. To show that a function is not one-to-one, find at least two x-values that produce the same y-values. Sketch the graph and use the horizontal line test. If the function either increases or decreases on its entire domain, then it is one-to-one. A sketch is helpful here, too.

9. Slide 4-9 Copyright © 2004 Pearson Education, Inc. Inverse Function Let f be one-to-one function. Then g is the inverse function of f if for every x in the domain of g, and for every x in the domain of f.

10. Slide 4-10 Copyright © 2004 Pearson Education, Inc. Deciding Whether Two Functions Are Inverses Example: Let functions f and g be defined by f(x) = 3x + 9 and g(x) =, respectively. Is g the inverse function of f ? Solution:The horizontal line test applied to the graph indicates that f is one-to-one. Since it is one-to-one, we now find: Since function g is not the inverse of function f.

11. Slide 4-11 Copyright © 2004 Pearson Education, Inc. Finding Inverses of One-to-One Functions Example: Find the inverse of the function if it is one-to-one. G = {(5,2), (3,1), (4,3), (2,0)} Solution: Every x-value in G corresponds to only one y-value., and every y-value corresponds to only one x-value, so G is a one-to-one function. The inverse function is found by interchanging the x- and y-values in each ordered pair. G  1 ={(2,5), (1,3), (3,4), (0,2)} Notice how the domain and range of G becomes the range and domain respectively, of G  1.

12. Slide 4-12 Copyright © 2004 Pearson Education, Inc. Equations of Inverses Finding the Equation of the Inverse of y = f(x). For a one-to-one function f defined by an equation y = f(x), find the defining equation of the inverse as follows. Step 1 Interchange x and y, Step 2 Solve for y. Step 3 Replace y with f  1 (x).

13. Slide 4-13 Copyright © 2004 Pearson Education, Inc. Finding Equations of Inverses f(x) = 3x + 7 Solution: The graph of y = 3x + 7 is a one-to-one function, verified by the horizontal line test. y = 3x + 7 x = 3y + 7 3y = x  7 y = f  1 (x) = y = x 2  5 Solution: The equation y = x 2  5 has a parabola opening down as its graph, so some horizontal lines will intersect the graph at two points. This function is not one-to-one and will not have an inverse.

14. Slide 4-14 Copyright © 2004 Pearson Education, Inc. Finding the Inverse of a Function with a Restricted Domain Example: Let f(x) = Find f –1 (x). Solution: First, notice that the domain of f is restricted to the interval [  3,  ). Function f is one-to-one because it is increasing on its entire domain, and thus has an inverse function.

15. Slide 4-15 Copyright © 2004 Pearson Education, Inc. Finding the Inverse of a Function with a Restricted Domain continued However, we cannot define f  1 (x) as x 2  3. The domain of f is [  3,  ), and its range is [0,  ). The range of f is the domain of f  1, so f  1 must be defined as f  1 (x) = x 2  3, x  0.

16. Slide 4-16 Copyright © 2004 Pearson Education, Inc. Inverses Important Facts about Inverses If f is one-to-one, then f  1 exists. The domain of f is equal to the range of f  1, and the range of f is equal to the domain of f  1. If the point (a, b) lies on the graph of f, then (b, a) lies on the graph of f  1,, so the graph of f and f  1 are reflections of each other across the line y = x. To find the equation for f  1, replace f(x) with y, interchange x and y, and solve for y. This gives f  1 (x).

17. Copyright © 2004 Pearson Education, Inc. 4.2 Exponential Functions

18. Slide 4-18 Copyright © 2004 Pearson Education, Inc. Additional Properties of Exponents If any real number a > 0, a  1, the following statements are true.  a x is a unique real number for all real numbers x.  a b = a c if and only if b = c.  If a > 1 and m < n, then a m < a n.  If 0 a n.

19. Slide 4-19 Copyright © 2004 Pearson Education, Inc. Evaluating an Exponential Expression If f(x) = 3 x, find each of the following. a) f(  1)b) f(3)c) f(3/2)d) f(5.01) Solutions: a) b) c) d)

20. Slide 4-20 Copyright © 2004 Pearson Education, Inc. Exponential Function

21. Slide 4-21 Copyright © 2004 Pearson Education, Inc. Exponential Function continued

22. Slide 4-22 Copyright © 2004 Pearson Education, Inc. Characteristics of the Graph of f(x) = a x 1. The points (0, 1), and (1, a) are on the graph. 2.If a > 1, then f is an increasing function; if 0 < a < 1, then f is a decreasing function. 3.The x-axis is a horizontal asymptote. 4.The domain is ( ,  ), and the range is (0,  ).

23. Slide 4-23 Copyright © 2004 Pearson Education, Inc. Example Graph f(x) = 6 x. y-intercept = 1 x-axis = horizontal asymptote Domain: ( ,  ) Range: (0,  ) 61 2.450.5 10 0.16 11 f(x)f(x)x

24. Slide 4-24 Copyright © 2004 Pearson Education, Inc. Graphing Reflections and Translations Graph each function. a) f(x) =  3 x b) f(x) = 3 x + 2 c) f(x) = 3 x + 2

25. Slide 4-25 Copyright © 2004 Pearson Education, Inc. Solution f(x) =  3 x Reflected across the x-axis. Domain: ( ,  ) Range: ( , 0)

26. Slide 4-26 Copyright © 2004 Pearson Education, Inc. Solution f(x) = 3 x + 2 The graph of f(x) = 3 x translated 2 units to the left.

27. Slide 4-27 Copyright © 2004 Pearson Education, Inc. Solution f(x) = 3 x + 2 The graph of f(x) = 3 x translated 2 units up.

28. Slide 4-28 Copyright © 2004 Pearson Education, Inc. Exponential Equations Solve

29. Slide 4-29 Copyright © 2004 Pearson Education, Inc. Another Example Solve 3 x + 1 = 27 x  3

30. Slide 4-30 Copyright © 2004 Pearson Education, Inc. Compound Interest If P dollars are deposited in an account paying an annual rate of interest r compounded (paid) m times per year, then after t years the account will contain A dollars, where

31. Slide 4-31 Copyright © 2004 Pearson Education, Inc. Example Suppose $2000 is deposited in an account paying 6% compounded semiannually (twice a year). a) Find the amount in the account after 10 years. b) How much interest is earned over the 10-yr period?

32. Slide 4-32 Copyright © 2004 Pearson Education, Inc. Solution a)  Thus, $3612.22 is an account after 10 yr. b) The interest earned for that period is $3612.22  $2000 = $1612.22

33. Slide 4-33 Copyright © 2004 Pearson Education, Inc. Definitions Value of e: To nine decimal places, e  2.718281828 Continuous Compounding If P dollars are deposited at a rate of interest r compounded continuously for t years, the compounded amount in dollars on deposit is A = Pe rt.

34. Slide 4-34 Copyright © 2004 Pearson Education, Inc. Example Suppose $2000 is deposited in an account paying 2% interest compounded continuously for 3 yr. Find the total amount on deposit at the end of the 3 yr. Solution:

35. Copyright © 2004 Pearson Education, Inc. 4.3 Logarithmic Functions

36. Slide 4-36 Copyright © 2004 Pearson Education, Inc. Logarithm For all real numbers y, and all positive numbers a and x, where a  1: Meaning of log a x A logarithm is an exponent; log a x is the exponent to which the base a must be raised to obtain x.

37. Slide 4-37 Copyright © 2004 Pearson Education, Inc. Solving Logarithmic Equations Solve each equation. a)b)

38. Slide 4-38 Copyright © 2004 Pearson Education, Inc. Logarithmic Function If a > 0, a  1, and x > 0, then defines the logarithmic function with base a. Logarithms can be found for positive numbers only.

39. Slide 4-39 Copyright © 2004 Pearson Education, Inc. Logarithmic Function

40. Slide 4-40 Copyright © 2004 Pearson Education, Inc. Logarithmic Function continued

41. Slide 4-41 Copyright © 2004 Pearson Education, Inc. Characteristics of the Graph of f(x) = log a x The points (1, 0), and (a, 1) are on the graph. If a > 1, then f is an increasing function; if 0 < a < 1, then f is a decreasing function. The y-axis is a vertical asymptote. The domain is (0,  ), and the range is ( ,  ).

42. Slide 4-42 Copyright © 2004 Pearson Education, Inc. Example Graph Write in exponential form as Now find some ordered pairs. 11 4 21/16 01 yx

43. Slide 4-43 Copyright © 2004 Pearson Education, Inc. Graph Write in exponential form as Now find some ordered pairs. Example 11 0.2 15 01 yx

44. Slide 4-44 Copyright © 2004 Pearson Education, Inc. Translated Logarithmic Functions Graph the function. The vertical asymptote is x = 1. To find some ordered pairs, use the equivalent exponent form.

45. Slide 4-45 Copyright © 2004 Pearson Education, Inc. Translated Logarithmic Functions continued Graph To find some ordered pairs, use the equivalent exponent form.

46. Slide 4-46 Copyright © 2004 Pearson Education, Inc. Properties of Logarithms, For x > 0, y > 0, a > 0, a  1, and any real number r: The logarithm of a number raised to a power is equal to the exponent multiplied by the logarithm of the number. Power Property The logarithm of the quotient of two numbers is equal to the difference between the logarithms of the numbers. Quotient Property The logarithm of a product of two numbers is equal to the sum of the logarithms of the numbers Product Property DescriptionProperty

47. Slide 4-47 Copyright © 2004 Pearson Education, Inc. Using the Properties of Logarithms Rewrite each expression. Assume all variables represent positive real numbers with a  1 and b  1. a) b) c)

48. Slide 4-48 Copyright © 2004 Pearson Education, Inc. Using the Properties of Logarithms Write each expression as a single logarithm with coefficient 1. Assume all variables represent positive real numbers with a  1 and b  1. a) b)

49. Slide 4-49 Copyright © 2004 Pearson Education, Inc. Theorem on Inverses For a > 0, a  1: By the results of this theorem:

50. Copyright © 2004 Pearson Education, Inc. 4.4 Evaluating Logarithms and the Change-of-Base Theorem

51. Slide 4-51 Copyright © 2004 Pearson Education, Inc. Common Logarithm For all positive numbers x, log x = log 10 x. A calculator with a log key can be used to find the base 10 logarithm of any positive number. In chemistry, the pH of a solution is defined as pH =  log[H 3 O + ], where [H 3 O + ] is the hydronium ion concentration in miles per liter.

52. Slide 4-52 Copyright © 2004 Pearson Education, Inc. Example Find the pH of a solution with [H 3 O + ] = 3.1  10  4 pH =  log[H 3 O + ] =  log(3.1  10  4 ) =  (log 3.1 + log 10  4 ) =  (.4914  4) = .4914 + 4 = 3.5086 The pH is  3.5.

53. Slide 4-53 Copyright © 2004 Pearson Education, Inc. Loudness of Sound The loudness of sounds is measured in a unit called a decibel. To measure with this unit, we first assign an intensity of I 0 to a very faint sound, called the threshold sound. If a particular sound has intensity I, then the decibel rating of this louder sound is

54. Slide 4-54 Copyright © 2004 Pearson Education, Inc. Example Find the decibel rating of a sound with intensity 1000I 0. The sound has a decibel rating of 30.

55. Slide 4-55 Copyright © 2004 Pearson Education, Inc. Natural Logarithm For all positive numbers x, ln x = log e x Natural logarithms can be found using a calculator.

56. Slide 4-56 Copyright © 2004 Pearson Education, Inc. Example Geologists sometimes measure the age of rocks by using “atomic clocks.” by measuring the amounts of potassium 40 and argon 40 in a rock, the age t of the specimen in years is found with the formula Where A and K are the numbers of atoms of argon 40 and potassium 40, respectively in the specimen.

57. Slide 4-57 Copyright © 2004 Pearson Education, Inc. Example continued How old is a rock in which A = 0 and K > 0? If A = 0 and A/K = 0 The rock is new (0 yr old). The ratio A/K for a sample of granite from New Hampshire is.212. How old is the sample? Since A/K =.212, we have The granite is about 1.85 billion yr old.

58. Slide 4-58 Copyright © 2004 Pearson Education, Inc. Change-of-Base Theorem For any positive real numbers x, a, and b, where a  1 and b  1:

59. Slide 4-59 Copyright © 2004 Pearson Education, Inc. Examples a) log 5 12b) log 2.4 Use the change-of-base theorem to find an approximation to four decimal places for each logarithm.

60. Copyright © 2004 Pearson Education, Inc. 4.5 Exponential and Logarithmic Equations

61. Slide 4-61 Copyright © 2004 Pearson Education, Inc. Properties of Logarithms If x > 0, y > 0, a > 0, and a  1, then x = y if and only if log a x = log a y.

62. Slide 4-62 Copyright © 2004 Pearson Education, Inc. Example Solve 8 x = 15 The solution set is {1.3023}.

63. Slide 4-63 Copyright © 2004 Pearson Education, Inc. Example Solvecontinued

64. Slide 4-64 Copyright © 2004 Pearson Education, Inc. Example Solve

65. Slide 4-65 Copyright © 2004 Pearson Education, Inc. Logarithmic Equations Solve The only valid solution is x = 4.

66. Slide 4-66 Copyright © 2004 Pearson Education, Inc. Example Solve:

67. Slide 4-67 Copyright © 2004 Pearson Education, Inc. Example Solve continued The solution set is {6, 2}.

68. Slide 4-68 Copyright © 2004 Pearson Education, Inc. Solving Exponential or Logarithmic Equations

69. Copyright © 2004 Pearson Education, Inc. 4.6 Applications and Models of Exponential Growth and Decay

70. Slide 4-70 Copyright © 2004 Pearson Education, Inc. The Exponential Growth or Decay Function In many situations that occur in ecology, biology, economics, and social sciences, a quantity changes at a rate proportional to the amount present. In such cases the amount present at time t is a special function of t called the exponential growth or decay function. Exponential Growth or Decay Function Let y 0 be the amount or number present at time t = 0. Then, under certain conditions, the amount present at any time t is modeled by y = y 0 e kt, where k is a constant.

71. Slide 4-71 Copyright © 2004 Pearson Education, Inc. Finding Doubling Time for Money Example: How long will it take for the money in an account that is compounded continuously at 4.5% interest to double? Solution: A = Pe rt 2P = Pe.045t 2 = e.045t ln 2 = ln e.045t ln 2 =.045t 15.4  t It will take about 15 years for the amount to double. Let A = 2P and r =.045

72. Slide 4-72 Copyright © 2004 Pearson Education, Inc. Determining an Exponential function to Model Population Growth Example: A certain city’s population reached 33 million people on January 14, 2003 and was growing at a rate of.0148. If this growth rate continues, the population in t years after January 15, 2003 will be given by the function defined by f(t) = 33e.0148t Based on this model, what will the city’s population be on January 14, 2009? In what year will the population reach 39 million?

73. Slide 4-73 Copyright © 2004 Pearson Education, Inc. Determining an Exponential function to Model Population Growth continued Solution: Since t = 0 on January 14, 2003, on January 14, 2009, t would be 2009  2003 = 6 years. We must find f(t) when t = 6. f(t) = 33e.0148t f(t) = 33e (.0148)6 Let t = 6.  36.1 According to the model, the population will be 36.1 million in 2009.

74. Slide 4-74 Copyright © 2004 Pearson Education, Inc. Determining an Exponential function to Model Population Growth continued Solution: The model predicts the city’s population would reach 39 million in the year 2003 + 11 = 2014.

75. Slide 4-75 Copyright © 2004 Pearson Education, Inc. Decay Function Models Example: If 900 g of a radioactive substance are present initially and 2 years later only 450 g remain, how much of the substance will be present after 8 years? Solution: To express the situation as an exponential equation, y = y 0 e kt, first find y 0 and then find k. 900 = y 0 e k(0) Let y = 900 and t = 0. 900 = y 0 e 0 = 1

76. Slide 4-76 Copyright © 2004 Pearson Education, Inc. Decay Function Models continued This gives y = 900e kt Substitute into y = y 0 e kt. 450 = 900e 2k Let y = 450 and t = 2. Divide by 900. Take the logarithms on both sides.

77. Slide 4-77 Copyright © 2004 Pearson Education, Inc. Decay Function Models continued Thus the exponential decay equation is y = 900e .347t. To find the amount present after 8 years, let t = 8. y = 900e .347(8)  900e  2.776  56 After 8 years, about 56 g of the substance will remain.

78. Slide 4-78 Copyright © 2004 Pearson Education, Inc. Modeling Newton’s Law of Cooling Newton’s law of cooling says that the rate at which a body cools is proportional to the difference C in temperature between the body and the environment around it. The temperature f(t) of the body at the time t in appropriate units after being introduced into an environment having constant temperature T 0 is f(t) = T 0 + Ce  kt, where C and k are constants. An object that is hotter than its surroundings will cool off, and an object that is cooler than its surroundings will warm up.

79. Slide 4-79 Copyright © 2004 Pearson Education, Inc. Example A gallon of orange juice with a temperature of 40  F is set down in a room with a temperature of 70  F. The juice warms to 50  F after 1 hour.  Write an equation to model the data.  Find the temperature of the orange juice after 2 hours.  After how many hours is the temperature of orange juice 65  F?

80. Slide 4-80 Copyright © 2004 Pearson Education, Inc. Solution We must find values for C and k. From the given information, when t = 0, T 0 = 70, and the temperature of the orange juice is f(0) = 40. Also when t = 1, f(1) = 50. Substitute the first pair of values in the equation along with T 0 = 70. f(t) = T 0 + Ce  kt Given formula 40 = 70 + Ce 0k Let t = 0, f(0) = 40, and T 0 = 70. 40 = 70 + C e 0 = 1  30 = C Subtract 70.

81. Slide 4-81 Copyright © 2004 Pearson Education, Inc. Solution continued Thus, now we use the remaining pair of values in this equation. f(t) = 70 + (  30)e -kt Let T 0 = 70 and C =  30 50 = 70  30e  1k Let t = 1 and f(1) = 50 20 = 30e  k Divide by 30. Take logarithms on both sides. ln e x = x

82. Slide 4-82 Copyright © 2004 Pearson Education, Inc. Solution continued Multiply by  1. Thus, the model is f(t) = 70  30e .405t. f(t) = 70  30e .405t Model from part (a). f(2) = 70  30e .405(2) Let t = 2  56.7  F 65 = 70  30e .405t Let f(t) = 65  5 =  30e .405t Subtract 70 5 = 30e .405t

83. Slide 4-83 Copyright © 2004 Pearson Education, Inc. Solution continued Divide by 30. Take the logarithms on both sides. ln e x = x t  4.42 The orange juice will warm to about 65  F in about 4.42 hours or about 4 hours 25 minutes.