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Logarithmic Equations Solving Logarithmic Equations

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8/15/2013 Logarithmic Equations 2 Solving Equations Solve 1. log 3 x = –3 With definition: Solution set: 27 1 = x With inverse function: 3 log 3 x = 3 –3 27 1 = x 3 –3 = x 27 1 { } log 3 x = –3 Exponentiate: Apply Definition:

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8/15/2013 Logarithmic Equations 3 Solving Equations Solve 2. = x 6 4 log 36 This is already solved for x, so simplify 6 1/4 6 2x = 4 1 = 2x 8 1 = x Solution set: { } 8 1 By definition the logarithm (i.e. x) is the power of the base that yields 6 4 6 4 36 x = WHY ?

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8/15/2013 Logarithmic Equations 4 Solving Equations Solve 3. x log x 5 = x4x4 = x 4 x4x4 4 5 = –x, for x < 0 By definition: log a x = b iffi a b = x x 4 = 5 5 = x 4 By inverse function: x 5 4 = Exponentiate: Apply Definition: x, for x > 0 = log x 5 = 4

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8/15/2013 Logarithmic Equations 5 Solving Equations Solve 3. log x 5 = 4 Solution set: { } 5 4 x 5 4 = 4 5 –x, for x < 0 x, for x > 0 = Question: Note: why doesn’t x 4 = 5 yield x = 5 4 ? then log x 5 is defined x 5 – = If with a negative base !! Since x 2 = 5 yields x = 5 2 then OR

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8/15/2013 Logarithmic Equations 6 Solving Equations Solve 4. 3 log 3 (x 2 + 5) By definition: ( 2 is the power of 3 that yields x 2 + 5 ) 3 2 = x 2 + 5 4 = x 2 { ± 2 } Solution set: = x 2 ± 2 = ± x log 3 (x 2 + 5) = 2 x 2 = 4 4 ± = x By inverse function: = 3 2 = x 2 + 5 9 9 = x 2 + 5 4 = x2x2 Exponentiate:

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8/15/2013 Logarithmic Equations 7 Solving Equations Solve 5. = x Question: Now, how do we find log 4 12 ? log 4 12 = log 4 (4 x ) To remove a variable from an exponent, find the logarithm of the exponential form 12 = 4 x

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8/15/2013 Logarithmic Equations 8 Solving Equations Remember: a x and log a x are inverses 1. To remove a variable from an exponent, find the logarithm of the exponential form 2. To remove a variable from a logarithm, exponentiate

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8/15/2013 Logarithmic Equations 9 One-to-One Property Review Since a x and log a x are 1-1 then 1. If x = y then a x = a y 2. If a x = a y then x = y 3. If x = y then log a x = log a y 4. If log a x = log a y then x = y These facts can be used to solve equations

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8/15/2013 Logarithmic Equations 10 Solving Equations: Examples 6. Solve Solution set: { 1, 3 } Apply zero product rule x 2x 2 7 = 7 4x – 3 Since the exponential function is 1-1 then x 2 = 4x – 3 x 2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x – 1 = 0 OR x – 3 = 0 x = 1ORx = 3

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8/15/2013 Logarithmic Equations 11 Solving Equations: Examples 6. Solve Solution set: { 1, 3 } x 2x 2 7 = 7 4x – 3 NOTE: x 2x 2 7 log 7 = 7 4x–3 log 7 Question: Could we use log or ln instead of log 7 ? x 2 = 4x – 3 x 2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 Could have applied inverse function x = 1, 3

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8/15/2013 Logarithmic Equations 12 Solving Equations: Examples 7. Solve 5 ln x = 10 By definition 2 is the power of the base that yields x Solution set: { ~ 7.38906 } x = e 2 ≈ (2.71828) 2 ≈ 7.38906 ln x = 2 Thus

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8/15/2013 Logarithmic Equations 13 Solving Equations: Examples 8. Solution set: { –2 } Solve log 3 (1 – x) = 1 By definition: Thus 3 1 = 1 – x x = 1 – 3 = –2

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8/15/2013 Logarithmic Equations 14 Solving Equations: Examples 9. Solve ln (x 2 – 4) – ln (x + 2) = ln (3 – x) ln x 2 – 4 x + 2 ( ) = ln (3 – x) x 2 – 4 x + 2 = 3 – x = 3x – x 2 + 6 – 2x Using the quotient rule Since the logarithm function is 1-1, then x 2 – 4 = (x + 2)(3 – x)

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8/15/2013 Logarithmic Equations 15 Solving Equations: Examples 9. Solve ln (x 2 – 4) – ln (x + 2) = ln (3 – x) = 3x – x 2 + 6 – 2x = (2x – 5)(x + 2) x 2 – 4 2x 2 – x – 10 = 0 2x – 5 = 0 x = 5/2 Question: What are the domain and range of the ln function ? Solution set: { 5/2 } Why not –2 ? ORx + 2 = 0 x = –2 OR

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8/15/2013 Logarithmic Equations 16 Quotient Example 10. Solve for x : Solving Equations log (x + 3) log (x + 1) = 2 log (x + 3) log (x + 1) = 2 = 2 x + 3 = (x + 1) 2 0 = x 2 + x – 2 = x 2 + 2x+ 1 = (x + 2)(x – 1) Solution set: { –2, 1 } WHY?

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8/15/2013 Logarithmic Equations 17 Think about it !

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