1. Slide 4-1 Copyright © 2005 Pearson Education, Inc.

2. Slide 4-2 Copyright © 2005 Pearson Education, Inc. Exponential and Logarithmic Functions Chapter 5

3. 5.1 Exponential Functions and Graphs  Graph exponential equations and functions.  Solve applied problems involving exponential functions and their graphs.

4. Slide 4-4 Copyright © 2005 Pearson Education, Inc. Exponential Function  The function f(x) = a x, where x is a real number, a > 0 and a  1, is called the exponential function, base a.  The base needs to be positive in order to avoid the complex numbers that would occur by taking even roots of negative numbers. Examples:

5. Slide 4-5 Copyright © 2005 Pearson Education, Inc. Graphing Exponential Functions  To graph an exponential function, follow the steps listed: 1. Compute some function values and list the results in a table. 2.Plot the points and connect them with a smooth curve. Be sure to plot enough points to determine how steeply the curve rises.

6. Slide 4-6 Copyright © 2005 Pearson Education, Inc. Example  Graph the exponential function y = f(x) = 3 x. (  3, 1/27) 1/27 33 (  2, 1/9) 1/9 22 (  1, 1/3) 1/3 11 (3, 27)273 9 3 1 y = f(x) = 3 x (2, 9)2 (1, 3)1 (0, 1)0 (x, y)x

7. Slide 4-7 Copyright © 2005 Pearson Education, Inc. Example  Graph the exponential function. (3, 1/27)1/273 (2, 1/9)1/92 (1, 1/3)1/31 (  3, 27) 27 33 9 3 1 (  2, 9) 22 (  1, 3) 11 (0, 1)0 (x, y)x

8. Slide 4-8 Copyright © 2005 Pearson Education, Inc. Example  Graph y = 3 x + 2. The graph is the graph of y = 3 x shifted to left 2 units. 2433 812 271 90 3 1 1/3 y 11 22 33 x

9. Slide 4-9 Copyright © 2005 Pearson Education, Inc. Example  Graph y = 4  3  x The graph is a reflection of the graph of y = 3 x across the x-axis, followed by a reflection across the y-axis and then a shift up of 4 units. 3.963 3.882 3.671 30 1 55  23 y 11 22 33 x

10. Slide 4-10 Copyright © 2005 Pearson Education, Inc. The Number e  e  2.7182818284…  Find each value of e x, to four decimal places, using the e x key on a calculator. a) e 4 b) e  0.25 c) e 2 d) e  1 a) 54.5982b) 0.7788c) 7.3891d) 0.3679

11. Slide 4-11 Copyright © 2005 Pearson Education, Inc. Graphs of Exponential Functions, Base e  Graph f(x) = e x. 7.3892 2.7181 1 0.368 0.135 f(x)f(x) 0 11 22 x

12. Slide 4-12 Copyright © 2005 Pearson Education, Inc. Example  Graph f(x) = 2  e  3x. 1.992 1.951 1  18.09  401.43 f(x)f(x) 0 11 22 x

13. Slide 4-13 Copyright © 2005 Pearson Education, Inc. Example  Graph f(x) = e x+2. 0.135 44 0.368 33 20.0861 7.389 2.718 1 f(x)f(x) 0 11 22 x

14. Logarithmic Functions and Graphs  Graph logarithmic functions.  Convert between exponential and logarithmic equations.  Find common and natural logarithms using a calculator.

15. Slide 4-15 Copyright © 2005 Pearson Education, Inc. Logarithmic Functions  These functions are inverses of exponential functions. Graph: x = 3 y. 1.Choose values for y. 2.Compute values for x. 3.Plot the points and connect them with a smooth curve. * Note that the curve does not touch or cross the y-axis.

16. Slide 4-16 Copyright © 2005 Pearson Education, Inc. Logarithmic Functions continued (1/27,  3) 33 1/27 (1/9,  2) 22 1/9 (1/3,  1) 11 1/3 2 1 0 y (9, 2)9 (3, 1)3 (1, 0)1 (x, y)x = 3 y Graph: x = 3 y

17. Slide 4-17 Copyright © 2005 Pearson Education, Inc. Logarithmic Function, Base a  We define y = log a x as that number y such that x = a y, where x > 0 and a is a positive constant other than 1.  We read log a x as “the logarithm, base a, of x.”

18. Slide 4-18 Copyright © 2005 Pearson Education, Inc. Finding Certain Logarithms  Find each of the following logarithms.  a) log 2 16b) log 10 1000  c) log 16 4d) log 10 0.001 a)The exponent to which we raise 2 to obtain 16 is 4; thus log 2 16 = 4. b) The exponent to which we raise 10 to obtain 1000 is 3; thus log 10 1000 = 3. c)The exponent we raise 16 to get 4 is ½, so log 16 4 = ½. d)We have The exponent to which we raise 10 to get 0.001 is  3, so log 10 0.001 =  3.

19. Slide 4-19 Copyright © 2005 Pearson Education, Inc. Logarithms  log a 1 = 0 and log a a = 1, for any logarithmic base a.   Convert each of the following to a logarithmic equation. a) 25 = 5 x b) e w = 30 log 5 25 = x log e 30 = w

20. Slide 4-20 Copyright © 2005 Pearson Education, Inc. Example  Convert each of the following to an exponential equation. a) log 7 343 = 3 log 7 343 = 3 7 3 = 343 b) log b R = 12 log b R = 12 b 12 = R The logarithm is the exponent. The base remains the same.

21. Slide 4-21 Copyright © 2005 Pearson Education, Inc. Example  Find each of the following common logarithms on a calculator. Round to four decimal places. a) log 723,456 b) log 0.0000245 c) log (  4) Does not exist ERR: nonreal ans log (  4)  4.6108  4.610833916 log 0.0000245 5.85945.859412123log 723,456 RoundedReadoutFunction Value

22. Slide 4-22 Copyright © 2005 Pearson Education, Inc. Natural Logarithms  Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus, ln xmeans log e x. ln 1 = 0 and ln e = 1, for the logarithmic base e.

23. Slide 4-23 Copyright © 2005 Pearson Education, Inc. Example  Find each of the following natural logarithms on a calculator. Round to four decimal places. a) ln 723,456 b) ln 0.0000245 c) ln (  4) Does not exist ERR: nonreal ans ln (  4)  10.6168  10.61683744 ln 0.0000245 13.491813.49179501ln 723,456 RoundedReadoutFunction Value

24. Slide 4-24 Copyright © 2005 Pearson Education, Inc. Changing Logarithmic Bases  The Change-of-Base Formula For any logarithmic bases a and b, and any positive number M,

25. Slide 4-25 Copyright © 2005 Pearson Education, Inc. Example Find log 6 8 using common logarithms. Solution: First, we let a = 10, b = 6, and M = 8. Then we substitute into the change-of-base formula:

26. Slide 4-26 Copyright © 2005 Pearson Education, Inc. Example  We can also use base e for a conversion. Find log 6 8 using natural logarithms. Solution: Substituting e for a, 6 for b and 8 for M, we have

27. Slide 4-27 Copyright © 2005 Pearson Education, Inc. Graphs of Logarithmic Functions  Graph: y = f(x) = log 6 x. Select y. Compute x. 22 1/36 11 1/6 3216 236 16 01 yx,or 6 y

28. Slide 4-28 Copyright © 2005 Pearson Education, Inc. Example  Graph each of the following. Describe how each graph can be obtained from the graph of y = ln x. Give the domain and the vertical asymptote of each function.  a) f(x) = ln (x  2)  b) f(x) = 2  ln x  c) f(x) = |ln (x + 1)|

29. Slide 4-29 Copyright © 2005 Pearson Education, Inc. Graph f(x) = ln (x  2)  The graph is a shift 2 units right. The domain is the set of all real numbers greater than 2. The line x = 2 is the vertical asymptote. 1.0995 0.6934 03  0.693 2.5  1.386 2.25 f(x)f(x)x

30. Slide 4-30 Copyright © 2005 Pearson Education, Inc. Graph f(x) = 2  ln x  The graph is a vertical shrinking, followed by a reflection across the x-axis, and then a translation up 2 units. The domain is the set of all positive real numbers. The y-axis is the vertical asymptote. 1.5985 1.7253 21 2.1730.5 2.5760.1 f(x)f(x)x

31. Slide 4-31 Copyright © 2005 Pearson Education, Inc. Graph f(x) = |ln (x + 1)|  The graph is a translation 1 unit to the left. Then the absolute value has the effect of reflecting negative outputs across the x-axis. The domain is the set of all real numbers greater than  1. The line x =  1 is the vertical asymptote. 1.9466 1.3863 0.6931 00  0.5 f(x)f(x)x

32. Slide 4-32 Copyright © 2005 Pearson Education, Inc. Application: Walking Speed  In a study by psychologists Bornstein and Bornstein, it was found that the average walking speed w, in feet per second, of a person living in a city of population P, in thousands, is given by the function w(P) = 0.37 ln P + 0.05.

33. Slide 4-33 Copyright © 2005 Pearson Education, Inc. Application: Walking Speed continued The population of Philadelphia, Pennsylvania, is 1,517,600. Find the average walking speed of people living in Philadelphia. Since 1,517,600 = 1517.6 thousand, we substitute 1517.6 for P, since P is in thousands: w(1517.6) = 0.37 ln 1517.6 + 0.05  2.8 ft/sec. The average walking speed of people living in Philadelphia is about 2.8 ft/sec.

34. Properties of Logarithmic Functions  Convert from logarithms of products, powers, and quotients to expressions in terms of individual logarithms, and conversely.  Simplify expressions of the type log a a x and.

35. Slide 4-35 Copyright © 2005 Pearson Education, Inc. Logarithms of Products  The Product Rule For any positive numbers M and N and any logarithmic base a, log a MN = log a M + log a N. (The logarithm of a product is the sum of the logarithms of the factors.)

36. Slide 4-36 Copyright © 2005 Pearson Education, Inc. Example  Express as a single logarithm:. Solution:

37. Slide 4-37 Copyright © 2005 Pearson Education, Inc. Logarithms of Powers  The Power Rule For any positive number M, any logarithmic base a, and any real number p, log a M p = p log a M. (The logarithm of a power of M is the exponent times the logarithm of M.)

38. Slide 4-38 Copyright © 2005 Pearson Education, Inc. Examples  Express as a product.

39. Slide 4-39 Copyright © 2005 Pearson Education, Inc. Logarithms of Quotients  The Quotient Rule For any positive numbers M and N, and any logarithmic base a,. (The logarithm of a quotient is the logarithm of the numerator minus the logarithm of the denominator.)

40. Slide 4-40 Copyright © 2005 Pearson Education, Inc. Examples  Express as a difference of logarithms.  Express as a single logarithm.

41. Slide 4-41 Copyright © 2005 Pearson Education, Inc. Applying the Properties  Express in terms of sums and differences of logarithms.

42. Slide 4-42 Copyright © 2005 Pearson Education, Inc. Example  Express as a single logarithm.

43. Slide 4-43 Copyright © 2005 Pearson Education, Inc. Final Properties  The Logarithm of a Base to a Power For any base a and any real number x, log a a x = x. (The logarithm, base a, of a to a power is the power.)  A Base to a Logarithmic Power For any base a and any positive real number x, (The number a raised to the power log a x is x.)

44. Slide 4-44 Copyright © 2005 Pearson Education, Inc. Examples  Simplify. a) log a a 6 b) ln e  8 Solution: a) log a a 6 = 6 b) ln e  8 =  8  Simplify. a) b) Solution: a) b)

45. 4.5 Solving Exponential and Logarithmic Equations  Solve exponential and logarithmic equations.

46. Slide 4-46 Copyright © 2005 Pearson Education, Inc. Solving Exponential Equations  Equations with variables in the exponents, such as 3 x = 40 and 5 3x = 25, are called exponential equations. Base-Exponent Property For any a > 0, a  1, a x = a y  x = y.

47. Slide 4-47 Copyright © 2005 Pearson Education, Inc. Example  Solve:. Write each side with the same base. Since the bases are the same number, 5, we can use the base- exponent property and set the exponents equal: Check: 5 2x  3 = 125 5 2(3)  3 ? 125 5 3 ? 125 125 = 125 True The solution is 3.

48. Slide 4-48 Copyright © 2005 Pearson Education, Inc. Another Property  Property of Logarithmic Equality For any M > 0, N > 0, a > 0, and a  1, log a M = log a N  M = N. Solve: 2 x = 50 This is an exact answer. We cannot simplify further, but we can approximate using a calculator. x  5.6439 We can check by finding 2 5.6439  50.

49. Slide 4-49 Copyright © 2005 Pearson Education, Inc. Example Solve: e  0.25w = 12 The solution is about  9.94.

50. Slide 4-50 Copyright © 2005 Pearson Education, Inc. Solving Logarithmic Equations  Equations containing variables in logarithmic expressions, such as log 2 x = 16 and log x + log (x + 4) = 1, are called logarithmic equations.  To solve logarithmic equations algebraically, we first try to obtain a single logarithmic expression on one side and then write an equivalent exponential equation.

51. Slide 4-51 Copyright © 2005 Pearson Education, Inc. Example  Solve: log 4 x =  3  Check: log 4 x =  3 The solution is

52. Slide 4-52 Copyright © 2005 Pearson Education, Inc. Example  Solve:  Check: For x = 3:  For x =  3: Negative numbers do not have real-number logarithms. The solution is 3.

53. Slide 4-53 Copyright © 2005 Pearson Education, Inc. Example  Solve: The value 6 checks and is the solution.

54. 4.6 Applications and Models: Growth and Decay  Solve applied problems involving exponential growth and decay.

55. Slide 4-55 Copyright © 2005 Pearson Education, Inc. Population Growth  The function P(t) = P 0 e kt, k > 0 can model many kinds of population growths. In this function: P 0 = population at time 0, P = population after time, t = amount of time, k = exponential growth rate. The growth rate unit must be the same as the time unit.

56. Slide 4-56 Copyright © 2005 Pearson Education, Inc. Example  Population Growth of the United States. In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau) Find the exponential growth function. What will the population be in 2020? After how long will the population be double what it was in 1990?

57. Slide 4-57 Copyright © 2005 Pearson Education, Inc. Solution  At t = 0 (1990), the population was about 249 million. We substitute 249 for P 0 and 0.08 for k to obtain the exponential growth function. P(t) = 249e 0.08t  In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t: P(3) = 249e 0.08(3) = 249e 0.24  317. The population will be approximately 317 million in 2020.

58. Slide 4-58 Copyright © 2005 Pearson Education, Inc. Solution continued  We are looking for the doubling time T. 498 = 249e 0.08T 2 = e 0.08T ln 2 = ln e 0.08T (Taking the natural logarithm on both sides) ln 2 = 0.08T (ln e x = x) = T 8.7  T The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077.

59. Slide 4-59 Copyright © 2005 Pearson Education, Inc. Interest Compound Continuously  The function P(t) = P 0 e kt can be used to calculate interest that is compounded continuously. In this function: P 0 = amount of money invested, P = balance of the account, t = years, k = interest rate compounded continuously.

60. Slide 4-60 Copyright © 2005 Pearson Education, Inc. Example  Suppose that $2000 is deposited into an IRA at an interest rate k, and grows to $5889.36 after 12 years. What is the interest rate? Find the exponential growth function. What will the balance be after the first 5 years? How long did it take the $2000 to double?

61. Slide 4-61 Copyright © 2005 Pearson Education, Inc. Solution  At t = 0, P(0) = P 0 = $2000. Thus the exponential growth function is P(t) = 2000e kt. We know that P(12) = $5889.36. We then substitute and solve for k: $5889.36 = 2000e 12k The interest rate is about 9%.

62. Slide 4-62 Copyright © 2005 Pearson Education, Inc. Solution continued  The exponential growth function is P(t) = 2000e 0.09t.  The balance after 5 years is P(5) = 2000e 0.09(5) = 2000e 0.45  $3136.62

63. Slide 4-63 Copyright © 2005 Pearson Education, Inc. Solution continued  To find the doubling time T, set P(T) = 2  P 0 = $4000 and solve for T. 4000 = 2000e 0.09T 2 = e 0.09T ln 2 = ln e 0.09T ln 2 = 0.09T = T 7.7  T The original investment of $2000 doubled in about 7.7 years.

64. Slide 4-64 Copyright © 2005 Pearson Education, Inc. Growth Rate and Doubling Time The growth rate k and the doubling time T are related by kT = ln 2 or or * The relationship between k and T does not depend on P 0.

65. Slide 4-65 Copyright © 2005 Pearson Education, Inc. Example  A certain town’s population is doubling every 37.4 years. What is the exponential growth rate? Solution:

66. Slide 4-66 Copyright © 2005 Pearson Education, Inc. Models of Limited Growth In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value. One model of such growth is which is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

67. Slide 4-67 Copyright © 2005 Pearson Education, Inc. Exponential Decay  Decay, or decline, of a population is represented by the function P(t) = P 0 e  kt, k > 0. In this function: P 0 = initial amount of the substance, P = amount of the substance left after time, t = time, k = decay rate.  The half-life is the amount of time it takes for half of an amount of substance to decay.

68. Slide 4-68 Copyright © 2005 Pearson Education, Inc. Graphs

69. Slide 4-69 Copyright © 2005 Pearson Education, Inc. Example Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined? Solution: We know (from Example 5 in our book), that the function for carbon dating is P(t) = P 0 e -0.00012t. If the charcoal has lost 7.3% of its carbon-14 from its initial amount P 0, then 92.7%P 0 is the amount present.

70. Slide 4-70 Copyright © 2005 Pearson Education, Inc. Example continued To find the age of the charcoal, we solve the equation for t : The charcoal was about 632 years old.