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171S5.5_p Exponential and Logarithmic Equations MAT 171 Precalculus Algebra Trigsted Pilot Test Dr. Claude Moore Cape Fear Community College CHAPTER 5: Exponential and Logarithmic Functions and Equations April 21, 2011 OBJECTIVES 1. Solving Exponential Equations 2. Solving Logarithmic Equations 5.1 Exponential Functions 5.2 The Natural Exponential Function 5.3 Logarithmic Functions 5.4 Properties of Logarithms 5.5 Exponential and Logarithmic Equations 5.6 Applications of Exponential and Logarithmic Functions Two Logarithmic properties If u = v, then logb u = logb v. Logarithm property of equality logb ur = r logb u. Power Rule for logarithms To Solve Exponential Equations 1. If the equation can be written in the form bu = bv, then solve the equation u = v 2. If the equation cannot easily be written in the form bu = bv. a. Use the logarithm property of equality to “take the log of both sides” (typically using the base 10 or e). b. Use the product rule of logarithms to “bring down” any exponents. c. Solve for the given variable. 1 171S5.5_p Exponential and Logarithmic Equations April 21, 2011 Properties of Logarithms If b > 0, b ≠ 1, u and v represent positive numbers and r is any real number, then Product rule: Solve each equation. Round to four decimals places. Quotient rule: Power rule: Logarithm Property of Equality (3x 2) (ln e) = ln 33 Solving Logarithmic Equations 1. Determine the Domain of the variable. If a logarithmic equation can be written in the form then u = v. Furthermore, if u = v, then Domain is solution to x – 6 > 0 or x > 6 Domain is solution to 5x+6 > 0 and 3x-2 > 0 x > -5/6 and x > 2/3 Which is x > 2/3 2. Use properties of logarithms to combine all logarithms, and write as a single logarithm, if needed. 3. Eliminate the logarithm by rewriting the equation in exponential form. 4. Solve for the given variable. 5. Check for any extraneous solutions. Verify that each solution is in the domain of the variable. Since 33/43 ≈ 0.767 is greater than 2/3 ≈ 0.667, x = 33/43 is the solution. 2 171S5.5_p Exponential and Logarithmic Equations April 21, 2011 Domain is solution to x – 2 > 0 or x > 2. 13. ex3 e3x+7 = 24 e(x3)+(3x+7) = 24 e4x+4 = 24 ln e4x+4 = ln 24 4x+4 = ln 24 4x = 4 + ln 24 x = 0.25(4 + ln 24) x ≈ 0.2055 14. 2(ex1)2 e3x = 80 2(e2x2) e3x = 80 e(2x2)+(3x) = 40 ex+1 = 40 ln ex+1 = ln 40 x+1 = ln 40 x = 1 + ln 40 x ≈ 2.6889 3 171S5.5_p Exponential and Logarithmic Equations April 21, 2011 2 4