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Thread: Log problem

  1. #1
    Member smmmc's Avatar
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    Log problem

    Solve for x

    loge(x)+loge(2x-1)=0
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  2. #2
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    Dear smmmc,

    I assume that you have written, $\displaystyle log_ex+log_e(2x-1)=0$

    In that case,

    $\displaystyle log_{e}x+log_{e}(2x-1)=0$

    $\displaystyle \ln{x}=-\ln{(2x-1)}$

    $\displaystyle \ln{x}=\ln(2x-1)^{-1}$

    $\displaystyle x=\frac{1}{2x-1}$

    $\displaystyle 2x^2-x-1=0$

    $\displaystyle x^2-\frac{1}{2}x-\frac{1}{2}=0$

    $\displaystyle (x-\frac{1}{4})^2-\frac{1}{16}-\frac{1}{2}=0$

    $\displaystyle (x-\frac{1}{4})^2=\frac{9}{16}$

    $\displaystyle x=\pm{\frac{3}{4}}+\frac{1}{4}$

    $\displaystyle x=1\mbox{ or }x=-\frac{1}{2}$

    Hope this helps.
    Last edited by Sudharaka; Jan 8th 2010 at 08:07 PM.
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  3. #3
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    Quote Originally Posted by smmmc View Post
    Solve for x

    loge(x)+loge(2x-1)=0
    $\displaystyle \ln{x} + \ln{(2x - 1)} = 0$

    $\displaystyle \ln{[x(2x - 1)]} = 0$

    $\displaystyle x(2x - 1) = e^0$

    $\displaystyle x(2x - 1) = 1$

    $\displaystyle 2x^2 - x = 1$

    $\displaystyle 2x^2 - x - 1 = 0$

    $\displaystyle 2x^2 - 2x + x - 1 = 0$

    $\displaystyle 2x(x - 1) + 1(x - 1) = 0$

    $\displaystyle (x - 1)(2x + 1) = 0$

    $\displaystyle x - 1 = 0$ or $\displaystyle 2x + 1 = 0$

    $\displaystyle x = 1$ or $\displaystyle x = -\frac{1}{2}$.
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  4. #4
    Member smmmc's Avatar
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    wow thanks! could you clarify what |n is?


    Thanks
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  5. #5
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    Quote Originally Posted by smmmc View Post
    wow thanks! could you clarify what |n is?


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    $\displaystyle \ln$ stands for natural logarithm, and is alternative notation for $\displaystyle \log_e$.
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  6. #6
    Member smmmc's Avatar
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    Thanks!
    Last edited by mr fantastic; Jan 8th 2010 at 10:08 PM. Reason: Moved a new question to a new thread.
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