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Thread: quick help with logs

  1. #1
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    quick help with logs

    solve:
    $\displaystyle \log_{7} \left ( {\frac{1}
    {\sqrt{7}}} \right )$

    how would you go about this w/o a calculator

    $\displaystyle 7^x = \frac{1}{\sqrt{7}}$

    or

    $\displaystyle \log_{7}1 - \log_{7}\sqrt{7} = x$
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by incyt3s View Post
    solve:
    $\displaystyle \log_{7} \left ( {\frac{1}
    {\sqrt{7}}} \right )$

    how would you go about this w/o a calculator

    $\displaystyle 7^x = \frac{1}{\sqrt{7}}$

    or

    $\displaystyle \log_{7}1 - \log_{7}\sqrt{7} = x$
    $\displaystyle \log_{7} \left( \frac{ 1 }{ \sqrt{7} } \right) = \log_{7} \left( 7^{- \frac{1}{2} } \right)$

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  3. #3
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    yess

    thanks
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  4. #4
    Super Member Aryth's Avatar
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    Quote Originally Posted by incyt3s View Post
    solve:
    $\displaystyle \log_{7} \left ( {\frac{1}
    {\sqrt{7}}} \right )$

    how would you go about this w/o a calculator

    $\displaystyle 7^x = \frac{1}{\sqrt{7}}$

    or

    $\displaystyle \log_{7}1 - \log_{7}\sqrt{7} = x$

    $\displaystyle \log_{7} \left ( {\frac{1}{\sqrt{7}}} \right ) = \log_{7}1 - \log_{7}\sqrt{7}$

    $\displaystyle \log_{7}1 = 0$

    $\displaystyle \log_{7} \left ( {\frac{1}{\sqrt{7}}} \right ) = - \log_{7}\sqrt{7}$

    $\displaystyle \log_{7}\sqrt{7} = \log_{7}7^\frac{1}{2} = \frac{1}{2}\log_{7}7$

    $\displaystyle \log_{7}7 = 1$

    $\displaystyle \log_{7} \left ( {\frac{1}{\sqrt{7}}} \right ) = - \frac{1}{2}$
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