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Thread: Logarithmic Equations

  1. #1
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    Logarithmic Equations

    Hi, so this gets me nowhere, especially after I get rid of the logarithm and I get 57/2. Maybe I'm missing something and you can point it out, this is what I have:

    log5x + (log5125/log5x) = 7/2
    log5x(125 - x) = 7/2
    log5(125x - x2) = 7/2
    125x - x2 = 57/2
    125x - x2 = 279.5
    x2 - 125x - 279.5 = 0

    Trial and error is no good, and quadratic seems to be the only option.

    Thanks
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  2. #2
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    Re: Logarithmic Equations

    Let $u=\log_5 x $

    Then $u+\dfrac{3}{u}=\dfrac{7}{2}$

    $2u^2-7u+6=0$

    $u=\dfrac{3}{2}$ or $u=2$

    $x=\sqrt{125}$ or $x=25$.
    Last edited by SlipEternal; Jun 11th 2017 at 07:34 PM.
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  3. #3
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    Re: Logarithmic Equations

    Quote Originally Posted by kaperdomo View Post
    Hi, so this gets me nowhere, especially after I get rid of the logarithm and I get 57/2. Maybe I'm missing something and you can point it out, this is what I have:
    log5x + (log5125/log5x) = 7/2
    It is painfully clear that you have no idea about this. Your question reduces to:
    $y+\dfrac{3}{y}=\dfrac{7}{2}$ where $y=\log_5(x)$ therefore $x=y^5$

    Can you solve for $y~?$
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  4. #4
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    Re: Logarithmic Equations

    Quote Originally Posted by Plato View Post

    Your question reduces to:
    $y+\dfrac{3}{y}=\dfrac{7}{2}$ where $y=\log_5(x)$ therefore $x=y^5$ . . . No.

    Can you solve for $y~?$
    Therefore  \  \ x = 5^y.
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