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Math Help - log challanges

  1. #1
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    Wink log challanges

    Need Help with This Question

    If Logba=c and logxb=c, show that logax=1/c^2
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  2. #2
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    Quote Originally Posted by requal View Post
    Need Help with This Question

    If Logba=c and logxb=c, show that logax=1/c^2

    Log_{b}(a) = c  \rightarrow a = b^c
    Log_{x}(a) = c Log_{x}(b)
    Log_{x}(a) = c^2
    \therefore Log_{a}(x) = \frac{1}{c^2}

    Bobak
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  3. #3
    MHF Contributor
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    Quote Originally Posted by requal View Post
    Need Help with This Question

    If Logba=c and logxb=c, show that logax=1/c^2
    Here is one way.

    log(b)[a] = c
    So,
    a = b^c
    Or,
    b = a^(1/c) ----(i)

    log(x)[b] = c
    So,
    b = x^c
    Or,
    x = b^(1/c) -----(ii)

    Plug the b from (i) into (ii),
    x = [a^(1/c)]^(1/c)
    x = a^(1 / c^2) -------(iii)

    Hence,
    log(a)[x]
    = log(a)[a^(1 / c^2)]
    = 1 / c^2
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