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Thread: Transposition of logs

  1. #1
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    Transposition of logs

    hi guys/girls

    im struggling with the transposition of the log to eventually find p1. im not sure how to write in this box as a equation so ive added it as a attachment to make it easier to read. i know lp and p2 from my question but need to find the value of p1.

    ive put it as far as i have got but i dont think its right. any help would be great

    Lp=20〖log〗_(10 ) (p2/p1)

    thanks in advance
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  2. #2
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    Re: Transposition of logs

    The inverse function to $\displaystyle \log_{10}x$ is $\displaystyle 10^x$. So

    $\displaystyle Lp=20\log_{10}\left(\frac{p_2}{p_1}\right)$

    implies

    $\displaystyle 10^{-Lp/20}}p_2=p_1$.

    Just in case, $\displaystyle 10^x=e^{x\ln10}$ where $\displaystyle \ln$ is logarithm to the base $\displaystyle e$.
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  3. #3
    MHF Contributor

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    Re: Transposition of logs

    First of all these are $\displaystyle \log_{10}$ not base $\displaystyle e$.
    So $\displaystyle p_1=p_2\cdot 10^{-\frac{Lp}{20}}.$
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