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Thread: [SOLVED] Weird Logarithm Question

  1. #1
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    [SOLVED] Weird Logarithm Question

    y^5.6 = 4

    Solve for y using logarithms.


    I know how to solve this question without logs, and I know how to solve logs when the variable is in the exponent.

    This is question is something that I never had to do before. Please help.
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  2. #2
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    You have to remember the rules for logs for this question:

    $\displaystyle y^{5.6} = 4 $
    $\displaystyle \ln y^{5.6} = \ln 4 $
    $\displaystyle 5.6\ln y = \ln 4 $
    $\displaystyle \ln y = \frac{\ln 4}{5.6} $
    $\displaystyle y = e^{\frac{\ln 4}{5.6}} $
    $\displaystyle y = 1.28.... $
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    Quote Originally Posted by Some_One View Post
    y^5.6 = 4

    Solve for y using logarithms.
    I'm assuming this is $\displaystyle y^{5.6}=4$ and that the period isn't used to represent multiplication.

    The solution depends on the use of a property of logarithms. I suggest you brush up on the various properties, but the one needed here is that $\displaystyle \log x^z=z\log x\text.$

    So, we start by taking the log of both sides, and then we can apply the property:

    $\displaystyle y^{5.6}=4\Rightarrow\ln y^{5.6}=\ln4$

    $\displaystyle \Rightarrow5.6\ln y=\ln4\Rightarrow\ln y=\frac{\ln4}{5.6}$

    Now, exponentiating both sides,

    $\displaystyle e^{\ln y}=e^{(\ln4)/5.6}\Rightarrow y=\left(e^{\ln4}\right)^{1/5.6}$

    $\displaystyle \Rightarrow y = 4^{1/5.6} = 4^{5/28} = 2^{5/14}\approx1.2809$

    I took the natural log (base $\displaystyle e$) of both sides, but you could use any base.
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