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Math Help - log proof

  1. #1
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    Red face log proof

    Prove that
    if \log_{x}y = am^{2} , for all a,b,m, bam = 1, then \log_{y}x = \frac{b}{m}



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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by hello View Post
    Prove that
    if \log_{x}y = am^{2} , for all a,b,m, bam = 1, then \log_{y}x = \frac{b}{m}



    note that am^2= am\cdot m = \frac{1}{b}\cdot m =\frac{m}{b}

    rewriting \log_{x}y=am^2 \iff x^{am^2}=y

    then using the above sub we get

    x^{\frac{m}{b}}=y \iff x=y^{\frac{b}{m}}

    rewriting in log form we get

    \log_{y}(x)=\frac{b}{m}
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  3. #3
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    Hello, hello!

    Are we allowed the "reciprocal theorem"? . \log_b(a) \:=\:\frac{1}{\log_a(b)}


    Given: . \log_{x}y = am^2,\:\text{ and for all } a,b,m\!:\;abm = 1

    Prove: . \log_{y}x = \frac{b}{m}

    We have: . abm = 1\quad\Rightarrow\quad a = \frac{1}{bm}

    Then: . \log_xy \:=\:am^2\:=\:\left(\frac{1}{bm}\right)m^2 \quad\Rightarrow\quad \log_xy \:=\:\frac{m}{b}

    The theorem says: . \log_xy \:=\:\frac{1}{\log_yx} \:=\:\frac{m}{b} \quad\Rightarrow\quad \log_yx \:=\:\frac{b}{m}

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