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Math Help - Integration Problem

  1. #1
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    Integration Problem

    Given

     \int \frac{\ln bx}{x^2} ~dx =-\frac{\ln bx}{x}-\frac{1}{x}+C

    Then find b where

     \int_{\frac{1}{b}}^4 \frac{5\ln bx}{x^2} ~dx = 5

    I'm saying that

     5\int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 5

    Dividing both sides by 5

     \int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 1

    \left[ -\frac{\ln bx}{x}-\frac{1}{x}\right]_{\frac{1}{b}}^4 = 1

    \left[ -\frac{\ln 4b}{4}-\frac{1}{4}\right]-\left[ -\frac{\ln b\frac{1}{b}}{\frac{1}{b}}-\frac{1}{\frac{1}{b}}\right] = 1

    \left[ -\frac{\ln 4b}{4}-\frac{1}{4}\right]-\left[ -0-b\right] = 1

     -\frac{\ln 4b}{4}-\frac{1}{4}+b = 1

    Here is where I am stuck because I can't find an explicit way to solve for b Am I missing a log law or something similar?

    Thanks..
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Everything is correct!

    You can't solve it for b.
    Last edited by Also sprach Zarathustra; August 5th 2010 at 06:18 PM.
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Mistake at the third line.

     \int_{\frac{1}{b}}^4 \frac{5\ln bx}{x^2} ~dx = 5

    I'm saying that

     5\int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 5

    Dividing both sides by 5

    I can't take a constant out?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    All this is after you fixing your mistake...
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  5. #5
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    I did an edit, sorry about that.

    Do you think I have solved the problem ok, in light of the edit?
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  6. #6
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    I suppose the question is, is the method correct?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    YES...
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