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Thread: How do I prove this Log?

  1. #1
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    How do I prove this Log?

    For a>1, b>1, show that (logab)(logba) = 1.

    The variable after the log is the base.

    How do I go about showing this?

    I already know these:

    Logbx = Logax / Logab

    (again, the variable after the log is the base, i don't know how to write it properly on this forum)

    and, Logtb = 1 / Logbt

    (variable after log is base)

    How do I approach / solve this problem?
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle\log_bx=\frac{\log_ax}{\log_ab}$.
    For $\displaystyle x=a$ the equality becomes
    $\displaystyle \log_ba=\frac{\log_aa}{\log_ab}=\frac{1}{\log_ab}$
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  3. #3
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    Okay...but how do I show that (logab)(logba) = 1?
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  4. #4
    MHF Contributor red_dog's Avatar
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    So, $\displaystyle \log_ba=\frac{1}{\log_ab}$
    Multiply the equality by $\displaystyle \log_ab$.
    You get $\displaystyle \log_ab\cdot\log_ba=1$
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  5. #5
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    $\displaystyle \log_b a = x$ and $\displaystyle \log_a b = y$.
    We want to show $\displaystyle xy=1$.
    But this means (by definition),
    $\displaystyle a=b^x \mbox{ and }a^y = b$.
    Thus,
    $\displaystyle a=b^x\implies (a^y)=(b^x)^y\implies a^y = b^{xy}\implies b =b^{xy}\implies xy=1$
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