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Thread: log proof

  1. #1
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    Red face log proof

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



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



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

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

    then using the above sub we get

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

    rewriting in log form we get

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

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


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

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

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

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

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

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