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Solving Exponential and Logarithmic Equations

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abbott
21 diapositives · 610 Views · Sep 10, 2014

Solving Exponential and Logarithmic Equations. JMerrill , 2005 Revised, 2008 LWebb , Rev. 2014. Same Base. Solve: 4 x-2 = 64 x 4 x-2 = (4 3 ) x 4 x-2 = 4 3x x–2 = 3x -2 = 2x -1 = x. 64 = 4 3.

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Solving Exponential and Logarithmic Equations
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  1. Solving Exponential and Logarithmic Equations JMerrill, 2005 Revised, 2008 LWebb, Rev. 2014

  2. Same Base • Solve: 4x-2 = 64x • 4x-2 = (43)x • 4x-2 = 43x • x–2 = 3x • -2 = 2x • -1 = x 64 = 43 If bM = bN, then M = N If the bases are already =, just solve the exponents

  3. You Do • Solve 27x+3 = 9x-1

  4. Review – Change Logs to Exponents • log3x = 2 • logx16 = 2 • log 1000 = x 32 = x, x = 9 x2 = 16, x = 4 10x = 1000, x = 3

  5. Using Properties to Solve Logarithmic Equations • 1. Condense both sides first (if necessary). • 2a. If there is only one side with a logarithm, then rewrite the equation in exponential form. • 2b. If the bases are the same on both sides, you can cancel the logs on both sides. • 3. Solve the simple equation.

  6. Solving by Rewriting as an Exponential • Solve log4(x+3) = 2 • 42 = x+3 • 16 = x+3 • 13 = x

  7. You Do • Solve 3ln(2x) = 12 • ln(2x) = 4 • Realize that our base is e, so • e4 = 2x • x ≈ 27.299 • You always need to check your answers because sometimes they don’t work!

  8. Example: Solve for x • log36 = log33 + log3x problem • log36 = log33x condense • 6 = 3x drop logs • 2 = x solution

  9. You Do: Solve for x • log 16 = x log 2 problem • log 16 = log 2xcondense • 16 = 2xdrop logs • x = 4 solution

  10. Example • 7xlog25 = 3xlog25 + ½ log225 • log257x = log253x + log225 ½ • log257x = log253x + log251 • log257x = log2(53x)(51) • log257x = log253x+1 • 7x = 3x + 1 • 4x = 1

  11. You Do: Solve for x • log4x = log44 problem • = log44 condense • = 4 drop logs • cube each side • X = 64 solution

  12. You Do • Solve: log77 + log72 = log7x + log7(5x – 3)

  13. You Do Answer • Solve: log77 + log72 = log7x + log7(5x – 3) • log714 = log7 x(5x – 3) • 14 = 5x2 -3x • 0 = 5x2 – 3x – 14 • 0 = (5x + 7)(x – 2) Do both answers work? NO!!

  14. Using Properties to Solve Logarithmic Equations • If the exponent is a variable, then take the natural log of both sides of the equation and use the appropriate property. • Then solve for the variable.

  15. Example: Solving • 2x = 7 problem • ln2x = ln7 take ln both sides • xln2 = ln7 power rule • x = divide to solve for x • x = 2.807

  16. Example: Solving • ex = 72 problem • lnex = ln 72 take ln both sides • x lne = ln 72 power rule • x = 4.277 solution: because ln e = ?

  17. You Do: Solving • 2ex + 8 = 20 problem • 2ex = 12 subtract 8 • ex = 6divide by 2 • ln ex = ln 6 take ln both sides • x lne = 1.792 power rule x = 1.792(remember: lne = 1)

  18. Example • Solve 5x-2 = 42x+3 • ln5x-2 = ln42x+3use log rules • (x-2)ln5 = (2x+3)ln4 distribute • xln5 – 2ln5 = 2xln4 + 3ln4  group x’s • xln5 – 2x ln4 = 3ln4 + 2ln5 factor GCF • x(ln5 – 2ln4) = 3ln4 + 2ln5  divide • x = 3ln4 + 2ln5 ln5 – 2ln4 • x≈-6.343

  19. You Do: • Solve 35x-2 = 8x+3 • ln 35x-2 = ln 8x+3 • (5x-2)ln3 = (x+3)ln8 • 5xln3 – 2ln3 = xln8 + 3ln8 • 5xln3 – xln8 = 3ln8 + 2ln3 • x(5ln3 – ln8) = 3ln8 + 2ln3 • x = 3ln8 + 2ln3 5ln3 – ln8 x = 2.471

  20. Final Example • How long will it take for $25,000 to grow to $500,000 at 9% annual interest compounded monthly?

  21. Example