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5.4 Exponential and Logarithmic Equations Essential Questions: How do we solve exponential and logarithmic equations? Solving Simple Equations Original Equation Rewritten Equation Solution a. 2 x 32 2 x 25 x5 b. ln xx ln3 0 ln x ln3 x 3 1 c. 9 3 d. e x 7 3 x 32 x 2 ln e x ln 7 x ln 7 e. ln x 3 eln x e3 f. log10 x 1 10 log10 x 10 1 x e3 x 101 1 10 Strategies for Solving Exponential and Logarithmic Equations 1. 2. 3. Rewrite the given equation in a form that allows the use of the One-to-One Properties of exponential or logarithmic functions. Rewrite an exponential equation in logarithmic form and apply the Inverse Property of logarithmic functions. Rewrite a logarithmic equation in exponential form and apply the Inverse Property of exponential functions. Solving Exponential Equations Solve each equation and approximate the result to three decimal places. x x a. e 72 b. 3(2 ) 42 Solution: e 72 ln e x ln 72 a. x x ln 72 x 4.277 b. Take natural log of each side. Use a calculator. 3(2 x ) 42 2 x 14 log 2 2 x log 2 14 x log 2 14 x 3.807 Solving an Exponential Equation Solve e 5 60 and approximate the result to three decimal places. x Solution: e 5 60 x e 55 x ln e x ln 55 Original equation. Subtract 5 from each side. Take natural log of each side. x ln 55 Inverse Property. x 4.007 Use a calculator. 2 n5 Solve decimal places. 2(3 2 n5 2(3 ) 4 11 ) 4 11 ) 15 15 2 n 5 3 2 2 n5 2(3 2 n5 log 3 3 15 log 3 2 and approximate the result to three 15 2n 5 log 3 2 2n 5 log3 7.5 5 1 n log3 7.5 2 2 n 3.417 Solving a Logarithmic Equation a. Solve ln x = 2 b. Solve log3 (5x 1) log3 ( x 7) Solution: a. ln x 2 eln x e 2 Exponentiate each side. x e2 b. log3 (5x 1) log3 ( x 7) 5x 1 x 7 4x 8 x2 Check this in the original equation. Solving a Logarithmic Equation Solve 5 2ln x 4 and approximate the result to three decimal places. Solution: 5 2ln x 4 2ln x 1 1 ln x 2 e ln x e 1 2 x e1 2 x 0.607 Exponentiate each side. Solving a Logarithmic Equation Solve 2log 5 3x 4. Solution: 2log5 3x 4 log5 3x 2 log5 3 x 5 5 2 3x 25 Exponentiate each side (base 5). 25 x 3 Checking for Extraneous Solutions Solve log10 5 x log10 ( x 1) 2 Solution: log10 5 x log10 ( x 1) 2 log10 [5 x( x 1)] 2 log10 (5 x2 5 x ) 10 10 2 5 x 5 x 100 2 5( x x 20) 0 2 ( x 4)( x 5) 0 x 4 x 5 The solutions appear to be 5 and -4. However, when you check these in the original equation, only x = 5 works.