Section 5.6 logarithmic and exponential equations
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This document contains solutions to several logarithmic and exponential equations. It begins by solving the equations log 4 2log x= and ( ) ( )2 2 1x x+ − =, finding the values of x that satisfy each. It then solves equations involving logarithms and exponents such as 3 7x = ln3 ln 7x, 5 2 3x × = 3, and 1 2 3 ln 2 ln5x x− + =. Each solution provides the step-by-step work and resulting value of x. The document concludes by solving the quadratic equation 9 3 6 0x x − − =.
Section 5.6 logarithmic and exponential equations
- 1. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Section 5.6 Logarithmic and Exponential Equations
- 2. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
- 3. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Recall:
- 4. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. 3 3Solve: log 4 2log x= Reminder: The domain of logarithmic functions is positive numbers only so check for extraneous solutions. 2 3 3log 4 log x= 2 4 x= 2 or 2x x= = −
- 5. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. ( ) ( )2 2Solve: log 2 log 1 1x x+ + − = ( ) ( )2log 2 1 1x x+ − = ( ) ( )2 2 1x x= + − 2 2 2x x= − − + 2 0x x+ = ( )1 0x x + = 0 or 1x x= = − Both solutions are in the domain of the log functions.
- 6. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
- 7. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Recall:
- 8. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 3 7x = ln3 ln 7x = ln3 ln 7x = ln 7 ln3 x = 1.771≈
- 9. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 5 2 3x × = 3 2 5 x = 3 ln 2 ln 5 x = ÷ 3 ln 2 ln 5 x = ÷ 3 ln 5 0.737 ln 2 x ÷ = ≈ −
- 10. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. 1 2 3 Solve: 2 5x x− + = 1 2 3 ln 2 ln5x x− + = ( ) ( )1 ln 2 2 3 ln5x x− = + ln 2 ln 2 2 ln5 3ln5x x− = + ln 2 2 ln5 3ln5 ln 2x x− = + ( )ln 2 2ln5 3ln5 ln 2x − = + 3ln5 ln 2 2.186 ln 2 2ln5 x + = ≈ − −
- 11. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solve: 9 3 6 0x x − − = ( ) 2 Note that 9 = 3 so the equation is quadratic in form.x x 2 3 and = 9x x u u= 2 6 0u u− − = ( ) ( )3 2 0u u− + = 3 or 2u u= = − 3 3 or 3 2x x = = − 1x = 3 0 for allx x>