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Math Help - Proving

  1. #1
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    Proving

    Show that if
    logb(a)=c and
    logy(b)=c, then
    loga(y)=c^-2
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  2. #2
    MHF Contributor
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    Hello KingV15
    Quote Originally Posted by KingV15 View Post
    Show that if
    logb(a)=c and
    logy(b)=c, then
    loga(y)=c^-2
    I generally reckon that the easiest way to solve log problems is to get rid of the logs as soon as possible! So let's try that here.

    \log_b(a)=c
    \Rightarrow a=b^c ...(1)
    \log_y(b)=c
    \Rightarrow b=y^c ...(2)
    and we need to show that:
    \log_a(y)=c^{-2}
    in other words that:
    y = a^{(c^{-2})}
    Now from (1):
    b = a^{(c^{-1})} ...(3)
    And from (2):
    y=b^{(c^{-1})}
    = \big(a^{(c^{-1}}\big)^{(c^{-1})} from (3)

     =a^{(c^{-2})}
    \Rightarrow \log_a(y)=c^{-2}
    Grandad
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  3. #3
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    Hello, KingV15!

    Good job, Grandad!
    You used my favorite approach.

    Here's another . . .



    Show that if: . \begin{array}{ccc}\log_b(a)\:=\:c & [1] \\ \log_y(b)\:=\:c & [2]\end{array}

    . . then: . \log_a(y)\:=\:c^{-2}

    Base-change Formula: . \log_b(N) \:=\:\frac{\log(N)}{\log(b)}


    \begin{array}{ccccccc}\text{From [1], we have:} & \dfrac{\log(a)}{\log(b)} \:=\:c & \Rightarrow & \dfrac{\log(b)}{\log(a)} \:=\:\dfrac{1}{c} & [3] \\ \\[-3mm]<br />
\text{From [2], we have:} & \dfrac{\log(b)}{\log(y)} \:=\:c & \Rightarrow & \dfrac{\log(y)}{\log(b)} \:=\:\dfrac{1}{c} & [4] \end{array}


    Multiply [3] by [4]: . \frac{\log(b)}{\log(a)}\cdot\frac{\log(y)}{\log(b)  } \:=\:\frac{1}{c}\cdot\frac{1}{c} \quad\Rightarrow\quad \frac{\log(y)}{\log(a)} \:=\:\frac{1}{c^2}


    . . Therefore: . \log_a(y) \:=\:c^{-2}

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