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Math Help - [SOLVED] Log identity problem

  1. #1
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    [SOLVED] Log identity problem

    Prove that :

    (Log(base 2)X -1)^ 2 / Log(base 2)X = Log(base 2)X + Log(base x)2 2

    I have tried but I always end up with Log(base 2)X + Log(base 2)X^-2 + 1

    All help appreciated
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  2. #2
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    We have to prove that \frac{\left(\log_2x-1\right)^2}{\log_2x}=\log_2x+\log_x2-2 ( x>0, x\ne 1).

    \frac{\left(\log_2x-1\right)^2}{\log_2x}=\log_2x+\frac{\log_22}{\log_2  x}-2\Leftrightarrow \frac{\left(\log_2x-1\right)^2}{\log_2x}=\log_2x+\frac{1}{\log_2x}-2\Leftrightarrow \left(\log_2x-1\right)^2=\log_2^2x+1-2\log_2x which is an identity. QED
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  3. #3
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    Hello, MarcoMP!

    We need this identity: . \log_ba \:=\:\frac{1}{\log_ab}


    Prove that: . \frac{(\log_2x - 1)^2}{\log_2x} \;=\;\log_2x + \log_x2 - 2

    On the left side we have:

    . . \frac{(\log_2x)^2 - 2\log_2x + 1}{\log_2x}

    . . =\;\frac{(\log_2x)^2}{\log_2x} - \frac{2\log_2x}{\log_2x} + \frac{1}{\log_2x}

    . . =\;\log_2x - 2 + \underbrace{\frac{1}{\log_2x}}_{\downarrow}
    . . = \;\log_2x - 2 + \log_x2

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