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Slide 10- 1 Copyright © 2012 Pearson Education, Inc.

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9.6 Solving Exponential and Logarithmic Equations ■ Solving Exponential Equations ■ Solving Logarithmic Equations

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Slide 9- 3 Copyright © 2012 Pearson Education, Inc. Solving Exponential Equations Equations with variables in exponents, such as 3 x = 5 and 7 3x = 90 are called exponential equations. In Section 12.3, we solved certain logarithmic equations by using the principle of exponential equality.

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Slide 9- 4 Copyright © 2012 Pearson Education, Inc. 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.)

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Slide 9- 5 Copyright © 2012 Pearson Education, Inc. Example Solution Solve: 5 x = 125. 5 x = 5 3 Note that 125 = 5 3. Thus we can write each side as a power of the same base: Since the base is the same, 5, the exponents must be equal. Thus, x must be 3. The solution is 3.

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Slide 9- 6 Copyright © 2012 Pearson Education, Inc. 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.)

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Slide 9- 7 Copyright © 2012 Pearson Education, Inc. Example Solution Solve: 3 x +1 = 43 We have The solution is (log 43/log 3) – 1, or approximately 2.4236. 3 x +1 = 43 log 3 x +1 = log 43 (x +1)log 3 = log 43 x +1 = log 43/log 3 x = (log 43/log 3) – 1 Principle of logarithmic equality Power rule for logs

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Slide 9- 8 Copyright © 2012 Pearson Education, Inc. continued We graph y 1 = 3 x +1 and y 2 = 43 Rounded to four decimal places, the x-coordinate of the point of intersection is 2.4236. The solution is 2.4236.

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Slide 9- 9 Copyright © 2012 Pearson Education, Inc. Example Solution Solve: e 1.32t = 2000 We have: Note that we use the natural logarithm Logarithmic and exponential functions are inverses of each other e 1.32t = 2000 ln e 1.32t = ln 2000 1.32t = ln 2000 t = (ln 2000)/1.32

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Slide 9- 10 Copyright © 2012 Pearson Education, Inc. continued We graph y 1 = e 1.32x and y 2 = 2000 Rounded to four decimal places, the solution is 5.7583.

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Slide 9- 11 Copyright © 2012 Pearson Education, Inc. To Solve an Equation of the Form a t = b for t 1.Take the logarithm (either natural or common) of both sides. 2.Use the power rule for exponents so that the variable is no longer written as an exponent. 3.Divide both sides by the coefficient of the variable to isolate the variable. 4.If appropriate, use a calculator to find an approximate solution in decimal form.

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Slide 9- 12 Copyright © 2012 Pearson Education, Inc. Solving Logarithmic Equations Equations containing logarithmic expressions are called logarithmic equations. We saw in Section 12.3 that certain logarithmic equations can be solved by writing an equivalent exponential equation.

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Slide 9- 13 Copyright © 2012 Pearson Education, Inc. Example Solution Solve: log 2 (6x + 5) = 4. 6x + 5 = 2 4 6x = 11 The solution is 11/6. The check is left to the student. log 2 (6x + 5) = 4 6x + 5 = 16 x = 11/6

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Slide 9- 14 Copyright © 2012 Pearson Education, Inc. 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.

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Slide 9- 15 Copyright © 2012 Pearson Education, Inc. Example Solution Solve: log x + log (x + 9) = 1. x 2 + 9x = 10 To increase the understanding, we write in the base 10. log 10 x + log 10 (x + 9) = 1 log 10 [x(x + 9)] = 1 x(x + 9) = 10 1 x 2 + 9x – 10 = 0 (x – 1)(x + 10) = 0 x – 1 = 0 or x + 10 = 0 x = 1 or x = –10

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Slide 9- 16 Copyright © 2012 Pearson Education, Inc. continued We graph y 1 = log(x) + log (x + 9) and y 2 = 1. There is one point of intersection (1, 1).

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Slide 9- 17 Copyright © 2012 Pearson Education, Inc. The logarithm of a negative number is undefined. The solution is 1. 0 + log (10) Check x = 1: log 1 + log (1 + 9) 0 + 1 = 1 TRUE x = –10: log (–10) + log (–10 + 9) FALSE

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Slide 9- 18 Copyright © 2012 Pearson Education, Inc. Example Solution Solve: log 3 (2x + 3) – log 3 (x – 1) = 2. log 3 (2x + 3) – log 3 (x – 1) = 2 log 3 [(2x + 3)/(x – 1)] = 2 (2x + 3)/(x – 1) = 3 2 (2x + 3)/(x – 1) = 9 (2x + 3) = 9(x – 1) x = 12/7 2x + 3 = 9x – 9 The solution is 12/7. The check is left to the student.

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