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Math Help - Proving the log equation

  1. #1
    Member smmmc's Avatar
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    Proving the log equation

    logb(a)+logc(b)+loga(c)=1/loga(b) + logb(c) + 1/logc(a)

    Hrmm.. im thinking loga(a)/loga(b), how does that equal to logb(a) ?


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  2. #2
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    Hello smmmc
    Quote Originally Posted by smmmc View Post
    logb(a)+logc(b)+loga(c)=1/loga(b) + logb(c) + 1/logc(a)

    Hrmm.. im thinking loga(a)/loga(b), how does that equal to logb(a) ?


    Thanks
    The definition of a logarithm is as follows:
    The log of a number, to a certain base, is that power to which the base must be raised to get the number.
    So, by definition, if
    x=\log_a(b)
    then
    a^x=b
    Therefore, using the laws of indices:
    a=b^{\frac1x}
    and so, using the definition of a log once again:
    \log_b(a)=\frac1x
    =\frac{1}{\log_a(b)}
    Grandad
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  3. #3
    Member smmmc's Avatar
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    i understand up to a=b^(1/x)

    but still dont get why it equals 1/loga(b)

    so from logb(a)=1/x, i just see this as b^(1/2)=a

    ?
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  4. #4
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    Dear smmmc,

    Think it this way,

    log_{a}b=x

    Grandad has shown, log_{b}a=\frac{1}{x}

    Therefore, by substitution log_{b}a=\frac{1}{log_{a}b}

    Hope this helps.
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  5. #5
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    Hello smmmc

    OK. I see that you get it to here:
    Quote Originally Posted by smmmc View Post
    i understand up to a=b^(1/x)
    But I'm not sure what you mean by this:
    ...so from logb(a)=1/x, i just see this as b^(1/2)=a
    I understand the difficulty you're having with logs, because the definition of a logarithm is a bit 'inside-out' somehow. But it's important for you to get your head around it, so here's a bit more explanation.

    If a=b^{\frac1x}, then \frac1x is the power to which b must be raised in order to give the answer a. Right?

    So, looking back to the definition of a log that I gave you at the beginning, this means that \frac1x = \log_b(a).

    Well if \log_b(a) =\frac1x and x = \log_a(b), then, by substitution, \log_b(a) = \frac{1}{\log_a(b)}.

    Grandad
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  6. #6
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    Quote Originally Posted by smmmc View Post
    i understand up to a=b^(1/x)

    but still dont get why it equals 1/loga(b)

    so from logb(a)=1/x, i just see this as b^(1/2)=a
    1/x, not 1/2. I assume that was a typo.
    ?
    Thanks
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