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Math Help - A logarithm prove question

  1. #1
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    A logarithm prove question

    Important * means multiply (a) means base a etc
    log(b)a*log(c)b*log(a)c=1
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  2. #2
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    Quote Originally Posted by kingkaisai2
    Important * means multiply (a) means base a etc
    log(b)a*log(c)b*log(a)c=1
    Hello,

    as you may know you can transform a logarithm, so you have only one base:

    \log_{b}{(a)}={\ln(a)\over\ln{(b)}

    Your problem can be written like this:

    {\ln(a)\over\ln{(b)}} \cdot {\ln(b)\over\ln{(c)}} \cdot {\ln(c)\over\ln{(a)}} =1

    Simplify the LHS of this equation and you'll get 1 = 1. Thus this equation is true for all \{a, b, c\}\in\mbox{]0;1[} \cup  \mbox{ ]1;\infty[}

    (Remark: The \cup means union of 2 sets of numbers. )
    Greetings

    EB
    Last edited by earboth; May 22nd 2006 at 11:47 AM.
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  3. #3
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    Hello, kingkaisai2!

    \log_b(a)\cdot\log_c(b)\cdot\log_a(c)\:=\:1
    If you don't know the Base-Change Formula,
    . . you'll have to do some Olympic-level gymnastics.


    Let \log_c(b) = P . . . then: c^P = b

    Take logs (base b): \log_b(c^P) = \log_b(b)\quad\Rightarrow\quad P\cdot\log_b(c) = 1

    Then: P = \frac{1}{\log_b(c)} . . . That is: \log_c(b) = \frac{1}{\log_b(c)}


    Let \log_a(c) = Q . . . then: a^Q = c

    Take logs (base b): \log_b(a^Q) = \log_b(c)\quad\Rightarrow\quad Q\cdot\log_b(a) = \log_b(c)

    Then: Q = \frac{\log_b(c)}{\log_b(a)} . . . That is: \log_a(c) = \frac{\log_b(c)}{\log_b(a)}



    So: \log_b(a)\cdot\log_c(b)\cdot\log_a(c) becomes:

    . . \log_b(a)\cdot\frac{1}{\log_b(c)}\cdot\frac{\log_b  (c)}{\log_b(a)} \;=\;1 . . . ta-DAA!
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