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Math Help - 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\log_bx=\frac{\log_ax}{\log_ab}.
    For x=a the equality becomes
    \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, \log_ba=\frac{1}{\log_ab}
    Multiply the equality by \log_ab.
    You get \log_ab\cdot\log_ba=1
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  5. #5
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    \log_b a = x and \log_a b = y.
    We want to show xy=1.
    But this means (by definition),
    a=b^x \mbox{ and }a^y = b.
    Thus,
    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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