CHAPTER 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions and Graphs 5.3 Logarithmic Functions and Graphs 5.4 Properties of Logarithmic Functions 5.5 Solving Exponential and Logarithmic Equations 5.6 Applications and Models: Growth and Decay; and Compound Interest Copyright © 2009 Pearson Education, Inc.

5.3 Logarithmic Functions and Graphs Find common logarithms and natural logarithms with and without a calculator. Convert between exponential and logarithmic equations. Change logarithmic bases. Graph logarithmic functions. Solve applied problems involving logarithmic functions. Copyright © 2009 Pearson Education, Inc.

Logarithmic Functions These functions are inverses of exponential functions. We can draw the graph of the inverse of an exponential function by interchanging x and y. To Graph: x = 2y. 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. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example Graph: x = 2y. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) This curve looks like the graph of y = 2x reflected across the line y = x, as we would expect for an inverse. The inverse of y = 2x is x = 2y. Copyright © 2009 Pearson Education, Inc.

Logarithmic Function, Base a We define y = loga x as that number y such that x = ay, where x > 0 and a is a positive constant other than 1. We read loga x as “the logarithm, base a, of x.” Copyright © 2009 Pearson Education, Inc.

Finding Certain Logarithms - Example Find each of the following logarithms. a) log10 10,000 b) log10 0.01 c) log2 8 d) log9 3 e) log6 1 f) log8 8 Solution: a) The exponent to which we raise 10 to obtain 10,000 is 4; thus log10 10,000 = 4. b) We have The exponent to which we raise 10 to get 0.01 is –2, so log10 0.01 = –2. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) c) log2 8: The exponent to which we raise 2 to get 8 is 3, so log2 8 = 3. d) log9 3: The exponent to which we raise 9 to get 3 is 1/2; thus log9 3 = 1/2. e) log6 1: 1 = 60. The exponent to which we raise 6 to get 1 is 0, so log6 1 = 0. f) log8 8: 8 = 81. The exponent to which we raise 8 to get 8 is 4, so log8 8 = 1. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Logarithms loga 1 = 0 and loga a = 1, for any logarithmic base a. A logarithm is an exponent! Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example Convert each of the following to a logarithmic equation. a) 16 = 2x b) 10–3 = 0.001 c) et = 70 The exponent is the logarithm. a) 16 = 2x log216 = x The base remains the same. Solution: b) 10–3 = 0.001 g log10 0.001 = –3 c) et = 70 g log e 70 = t Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example Convert each of the following to an exponential equation. a) log 2 32= 5 b) log a Q= 8 c) x = log t M The logarithm is the exponent. a) log 2 32 = 5 25 = 32 The base remains the same. Solution: b) log a Q = 8 g a8 = Q c) x = log t M g tx = M Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example Find each of the following common logarithms on a calculator. Round to four decimal places. a) log 645,778 b) log 0.0000239 c) log (3) Solution: Function Value Readout Rounded a) log 645,778 5.8101 b) log 0.0000239 –4.6216 c) log (–3) Does not exist. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Natural Logarithms Logarithms, base e, are called natural logarithms. The abbreviation “ln” is generally used for natural logarithms. Thus, ln x means loge x. ln 1 = 0 and ln e = 1, for the logarithmic base e. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example Find each of the following natural logarithms on a calculator. Round to four decimal places. a) ln 645,778 b) ln 0.0000239 c) log (5) d) ln e e) ln 1 Solution: Function Value Readout Rounded a) ln 645,778 13.3782 b) ln 0.0000239 –10.6416 Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: Function Value Readout Rounded c) ln (–5) Does not exist. b) ln e 1 c) ln 1 0 Copyright © 2009 Pearson Education, Inc.

Changing Logarithmic Bases The Change-of-Base Formula For any logarithmic bases a and b, and any positive number M, Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example Find log5 8 using common logarithms. Solution: First, we let a = 10, b = 5, and M = 8. Then we substitute into the change-of-base formula: Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example We can also use base e for a conversion. Find log5 8 using natural logarithms. Solution: Substituting e for a, 6 for b and 8 for M, we have Copyright © 2009 Pearson Education, Inc.

Graphs of Logarithmic Functions - Example Graph: y = f (x) = log5 x. Solution: Method 1 y = log5 x is equivalent to x = 5y. Select y and compute x. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) Graph: y = f (x) = log5 x. Solution: Method 2 Use a graphing calculator. First change bases. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) Graph: y = f (x) = log5 x. Solution: Method 3 Calculators which graph inverses automatically. Begin with y1 = 5x, the graphs of both y1 and its inverse y2 = log 5 x will be drawn. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 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 + 3) b) f (x) = 3  ln x c) f (x) = |ln (x – 1)| Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) a) f (x) = ln (x + 3) The graph is a shift 3 units left. The domain is the set of all real numbers greater than –3, (–3, ∞). The line x = –3 is the vertical asymptote. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) b) f (x) = 3 – ln x The graph is a vertical shrinking of y = ln x, followed by a reflection across the x-axis and a translation up 3 units. The domain is the set of all positive real numbers, (0, ∞). The y-axis is the vertical asymptote. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) c) f (x) = |ln (x – 1)| The graph is a translation of y = ln x, right 1 unit. The effect of the absolute is to reflect the negative output across the x-axis. The domain is the set of all positive real numbers greater than 1, (1, ∞). The line x =1 is the vertical asymptote. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Application 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. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example a. The population of Hartford, Connecticut, is 124,848. Find the average walking speed of people living in Hartford. The population of San Antonio, Texas, is 1,236,249. Find the average walking speed of people living in San Antonio. Graph the function. d. A sociologist computes the average walking speed in a city to the approximately 2.0 ft/sec. Use this information to estimate the population of the city. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) Solution: a. Since P is in thousands and 124,848 = 124.848 thousand, we substitute 124.848 for P: w(124.848) = 0.37 ln 124.848 + 0.05  1.8 ft/sec. The average walking speed of people living in Hartford is about 1.8 ft/sec. b. Substitute 1236.249 for P: w(1236.249) = 0.37 ln 1236.249 + 0.05  2.7 ft/sec. The average walking speed of people living in San Antonio is about 2.7 ft/sec. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) c. Graph with a viewing window [0, 600, 0, 4] because inputs are very large and outputs are very small by comparison. Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc. Example (continued) d. To find the population for which the walking speed is 2.0 ft/sec, we substitute 2.0 for w(P), 2.0 = 0.37 ln P + 0.05, and solve for P. Use the Intersect method. Graph y1 = 0.37 ln x + 0.05 and y2 = 2. In a city with an average walking speed of 2.0 ft/sec, the population is about 194.5 thousand or 194,500. Copyright © 2009 Pearson Education, Inc.