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Math Help - Contrived Log problem

  1. #1
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    Contrived Log problem

    Solve for x: \log_{4}x = \sqrt{\log_{4}x}


    Thanks.
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by Savior_Self View Post
    Solve for x: \log_{4}x = \sqrt{\log_{4}x}


    Thanks.
    \log _4 x = \sqrt{\log _4 x }

    \log _4 x - \sqrt{\log _4 x } = 0

    \sqrt{\log _4 x }(\sqrt{\log _4 x} -1)=0

    \sqrt{\log _4 x} =0 \Rightarrow x = 4^0 =1

    \sqrt{\log _4 x } -1 =0

    \sqrt{\log _4 x } =1

    \log _4 x = 1 \Rightarrow x =4^1 = 4
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  3. #3
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    Answer is x = e^(ln(4)) = 4.

    Suppose equitation x=sqrt(x), it can be only if x=1,
    So we have log_4(x)=1 => x=4
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