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Thread: Solve for x without log's

  1. #1
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    Solve for x without log's

    Solve for x $\displaystyle ln\, x=1/5 ln(A-1) - 4 ln\,B +4$, expressed without the log's.

    My attempt:
    $\displaystyle
    ln\, x = \frac{{1}}{5}ln\,(A-1)-4ln\, B+4 \\ $
    $\displaystyle ln\, x = ln\,(A-1)^{1/5}-ln\, B^{4}+4 \\$
    $\displaystyle ln\, x = ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4 \\$
    $\displaystyle e^{ln\, x} = e^{ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4} \\$
    $\displaystyle x = \left(\frac{(A-1)^{1/5}}{B^{4}}\right)\times e^{4}$

    Can this be done better; or condensed more?

    Thanks!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Jodles View Post
    Solve for x $\displaystyle ln\, x=1/5 ln(A-1) - 4 ln\,B +4$, expressed without the log's.

    My attempt:
    $\displaystyle
    ln\, x = \frac{{1}}{5}ln\,(A-1)-4ln\, B+4 \\ $
    $\displaystyle ln\, x = ln\,(A-1)^{1/5}-ln\, B^{4}+4 \\$
    $\displaystyle ln\, x = ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4 \\$
    $\displaystyle e^{ln\, x} = e^{ln\,\left(\frac{(A-1)^{1/5}}{B^{4}}\right)+4} \\$
    $\displaystyle x = \left(\frac{(A-1)^{1/5}}{B^{4}}\right)\times e^{4}$

    Can this be done better; or condensed more?

    Thanks!
    From your second line if you say that $\displaystyle 4 = \ln(e^4)$ then your second line becomes

    $\displaystyle \ln(x) = \ln \left(\dfrac{\sqrt[5]{A-1} \cdot e^4}{B^4}\right)$

    The continue as normal to get the same answer.

    Your way is fine though, if anything quicker
    Last edited by e^(i*pi); Nov 14th 2010 at 08:55 AM. Reason: latex
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