Copyright © 2006 Pearson Education, Inc Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Exponential and Logarithmic Functions 9 Exponential and Logarithmic Functions 9.1 Composite and Inverse Functions 9.2 Exponential Functions 9.3 Logarithmic Functions 9.4 Properties of Logarithmic Functions 9.5 Common and Natural Logarithms 9.6 Solving Exponential and Logarithmic Equations 9.7 Applications of Exponential and Logarithmic Functions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Exponential and Logarithmic Equations 9.6 Solving Exponential and Logarithmic Equations Solving Exponential Equations Solving Logarithmic Equations Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Exponential Equations Equations with variables in exponents, such as 3x = 5 and 73x = 90 are called exponential equations. In Section 9.3, we solved certain exponential equations by using the principle of exponential equality. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Exponential Equality For any real number b, where (Powers of the same base are equal if and only if the exponents are equal.) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: 5x = 125. Note that 125 = 53. Thus we can write each side as a power of the same base: 5x = 53 Since the base is the same, 5, the exponents must be equal. Thus, x must be 3. The solution is 3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Logarithmic Equality For any logarithmic base a, and for x, y > 0, (Two expressions are equal if and only if the logarithms of those expressions are equal.) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: 3 x +1 = 43 We have 3 x +1 = 43 Principle of logarithmic equality log 3 x +1 = log 43 (x +1)log 3 = log 43 Power rule for logs x +1 = log 43/log 3 x = (log 43/log 3) – 1 The solution is (log 43/log 3) – 1, or approximately 2.4236. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: e1.32t = 2000 We have: e1.32t = 2000 Note that we use the natural logarithm ln e1.32t = ln 2000 Logarithmic and exponential functions are inverses of each other 1.32t = ln 2000 t = (ln 2000)/1.32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

To Solve an Equation of the Form at = b for t Take the logarithm (either natural or common) of both sides. Use the power rule for exponents so that the variable is no longer written as an exponent. Divide both sides by the coefficient of the variable to isolate the variable. If appropriate, use a calculator to find an approximate solution in decimal form. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Logarithmic Equations Equations containing logarithmic expressions are called logarithmic equations. We saw in Section 9.3 that certain logarithmic equations can be solved by writing an equivalent exponential equation. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: log2(6x + 5) = 4. log2(6x + 5) = 4 6x + 5 = 24 The solution is 11/6. The check is left to the student. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Often the properties for logarithms are needed Often the properties for logarithms are needed. The goal is to first write an equivalent equation in which the variable appears in just one logarithmic expression. We then isolate that expression and solve as in the previous example. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: log x + log (x + 9) = 1. log[x(x + 9)] = 1 x(x + 9) = 101 x2 + 9x = 10 x2 + 9x – 10 = 0 (x – 1)(x + 10) = 0 x – 1 = 0 or x + 10 = 0 x = 1 or x = –10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Check x = 1: log 1 + log (1 + 9) 0 + log (10) 0 + 1 = 1 x = –10: 0 + 1 = 1 TRUE x = –10: log (–10) + log (–10 + 9) FALSE The logarithm of a negative number is undefined. The solution is 1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solution Solve: log3(2x + 3) – log3(x – 1) = 2. log3[(2x + 3)/(x – 1)] = 2 (2x + 3)/(x – 1) = 32 (2x + 3)/(x – 1) = 9 (2x + 3) = 9(x – 1) 2x + 3 = 9x – 9 x = 12/7 The solution is 12/7. The check is left to the student. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley