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Solving Exponential and Logarithmic Equations in Algebra 2

Learn to solve exponential and logarithmic equations using the Change of Base Formula and algebraic techniques in this Algebra 2 lesson. Practice problems included.

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Solving Exponential and Logarithmic Equations in Algebra 2

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  1. Ch. 8.5 Exponential and Logarithmic Equations

  2. log 16 2 log 5 x = Divide each side by 2 log 5. 0.8614 Use a calculator. Check: 52x 16 52(0.8614) 16 Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 52x = 16. 52x = 16 log 52x = log 16 Take the common logarithm of each side. 2x log 5 = log 16 Use the power property of logarithms. 8-5

  3. Check understanding 1

  4. Change of Base Formula For any positive numbers, M, b, and c, with b ≠ 1, and c ≠ 1,

  5. log 12 log 6 log6 12 = Use the Change of Base Formula. 1.0792 0.7782 1.387 Use a calculator. log6 12 = log3xWrite an equation. 1.387 log3xSubstitute log6 12 = 1.3868 log x log 3 1.387 Use the Change of Base Formula. Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Use the Change of Base Formula to evaluate log6 12. Then convert log6 12 to a logarithm in base 3. 8-5

  6. 1.387 • log 3 log xMultiply each side by log 3. 1.387 • 0.4771 log xUse a calculator. 0.6617 log xSimplify. x 100.6617Write in exponential form. 4.589 Use a calculator. Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 (continued) The expression log6 12 is approximately equal to 1.3869, or log3 4.589. 8-5

  7. Check understanding 2

  8. log 120 log 5 2x = Use the Change of Base Formula. x 1.487 Use a calculator to solve for x. Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 52x = 120. 52x = 120 log5 52x = log5 120 Take the base-5 logarithm of each side. 2x = log5 120 Simplify. 8-5

  9. Check understanding 3

  10. Graph the equations y = 43x and y = 1100. Find the point of intersection. The solution is x 1.684 Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 43x = 1100 by graphing. 8-5

  11. Check: log (2x– 2) 4 log (2 •5001– 2) 4 log 10,000 4 log 104 = 4 Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve log (2x– 2) = 4. log (2x– 2) = 4 2x– 2 = 104 Write in exponential form. 2x– 2 = 10000 x = 5001 Solve for x. 8-5

  12. x3 2 Log ( ) = 5 Write as a single logarithm. x3 2 = 105 Write in exponential form. 3 x = 10 200, or about 58.48. 3 The solution is 10 200, or about 58.48. Exponential and Logarithmic Equations ALGEBRA 2 LESSON 8-5 Solve 3 log x– log 2 = 5. 3 log x– log 2 = 5 x3 = 2(100,000) Multiply each side by 2. 8-5

  13. Check understanding 6

  14. Homework P. 464 # 2- 22 even

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