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CHAPTER 5: Exponential and Logarithmic Functions 5.1 5.2 5.3 5.4 5.5 Inverse Functions Exponential Functions and Graphs Logarithmic Functions and Graphs Properties of Logarithmic Functions Solving Exponential and Logarithmic Equations 5.6 Applications and Models: Growth and Decay; and Compound Interest Copyright © 2009 Pearson Education, Inc. 5.5 Solving Exponential and Logarithmic Equations Solve exponential equations. Solve logarithmic equations. Copyright © 2009 Pearson Education, Inc. Solving Exponential Equations Equations with variables in the exponents, such as 3x = 20 and 25x = 64, are called exponential equations. Use the following property to solve exponential equations. Base-Exponent Property For any a > 0, a 1, ax = ay x = y. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 4 Example Solve 2 3x7 32. Solution: Write each side as a power of the same number (base). 2 3x7 25 Since the bases are the same number, 2, we can use the base-exponent property and set the exponents equal: 3x 7 5 3x 12 Check x = 4: x4 The solution is 4. Copyright © 2009 Pearson Education, Inc. 2 3x7 32. 2 34 7 ? 32 2127 25 32 32 TRUE Slide 5.5 - 5 Another Property Property of Logarithmic Equality For any M > 0, N > 0, a > 0, and a 1, loga M = loga N M = N. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 6 Example Solve: 3x = 20. Solution: 3x 20 log 3 log20 x x log 3 log 20 This is an exact answer. We cannot simplify further, but we can approximate using a calculator. log 20 x 2.7268 log 3 We can check by finding 32.7268 20. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 7 Example Solve: e0.08t = 2500. Solution: e0.08t 2500 lne0.08t ln2500 0.08t ln(2500) ln(2500) t 0.08 t 97.8 The solution is about 97.8. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 8 Solving Logarithmic Equations Equations containing variables in logarithmic expressions, such as log2 x = 4 and log x + log (x + 3) = 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. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 9 Example Solve: log3 x = 2. Solution: log 3 x 2 32 x 1 x 2 3 1 x 9 Copyright © 2009 Pearson Education, Inc. 1 Check: x 9 log 3 x 2 1 log 3 ? 2 9 log 3 32 2 2 TRUE 1 The solution is . 9 Slide 5.5 - 10 Example Solve: log x log x 3 1. Solution: log x log x 3 1 log x x 3 1 x x 3 101 x 2 3x 10 x 2 3x 10 0 x 2 x 5 0 x 2 0 or x 5 0 x 2 or x 5 Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 11 Example (continued) Check x = –5: Check x = 2: log x log x 3 1 log 2 log 2 3 ? 1 log 2 log 5 log 2 5 log10 1 log x log x 3 1 log 5 log 5 3 ? 1 FALSE 1 TRUE The number –5 is not a solution because negative numbers do not have real number logarithms. The solution is 2. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 12 Example Solve: ln 4x 6 ln x 5 ln x. Solution: ln 4x 6 ln x 5 ln x 4x 6 ln ln x x5 4x 6 x x5 4x 6 x x 5 x 5 x5 4x 6 x 2 5x Copyright © 2009 Pearson Education, Inc. 0 x2 x 6 0 x 3x 2 x 3 0 or x 2 0 x 3 or x 2 Only the value 2 checks and it is the only solution. Slide 5.5 - 13 Example - Using the Graphing Calculator Solve: e0.5x – 7.3 = 2.08x + 6.2. Solve: Graph y1 = e0.5x – 7.3 and y2 = 2.08x + 6.2 and use the Intersect method. The approximate solutions are –6.471 and 6.610. Copyright © 2009 Pearson Education, Inc. Slide 5.5 - 14