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Section 5.5 Solving Exponential and Logarithmic Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

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Presentation on theme: "Section 5.5 Solving Exponential and Logarithmic Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc."— Presentation transcript:

1 Section 5.5 Solving Exponential and Logarithmic Equations Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

2 Objectives  Solve exponential equations.  Solve logarithmic equations.

3 Solving Exponential Equations Equations with variables in the exponents, such as 3 x = 20 and 2 5x = 64, are called exponential equations. Use the following property to solve exponential equations. Base-Exponent Property For any a > 0, a  1, a x = a y  x = y.

4 Example Solve: Write each side as a power of the same number (base). Since the bases are the same number, 2, we can use the base-exponent property and set the exponents equal: Check x = 4: TRUE The solution is 4.

5 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.

6 Example Solve: 3 x = 20. This is an exact answer. We cannot simplify further, but we can approximate using a calculator. We can check by finding 3 2.7268  20.

7 Example Solve: 100e 0.08t = 2500. The solution is about 40.2.

8 Example Solve: 4 x+3 = 3 -x.

9 Solving Logarithmic Equations Equations containing variables in logarithmic expressions, such as log 2 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.

10 Example Solve: log 3 x =  2. The solution is TRUE Check:

11 Example Solve:

12 Example (continued) Check x = –5: FALSE Check x = 2: TRUE The number –5 is not a solution because negative numbers do not have real number logarithms. The solution is 2.

13 Example Solve: Only the value 2 checks and it is the only solution.

14 Example - Using the Graphing Calculator Solve: e 0.5x – 7.3 = 2.08x + 6.2. Graph y 1 = e 0.5x – 7.3 and y 2 = 2.08x + 6.2 and use the Intersect method. The approximate solutions are –6.471 and 6.610.


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