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Math Help - Need Help Solving a Proof for a logarithmic function

  1. #1
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    Need Help Solving a Proof for a logarithmic function

    Hello, I'm currently stuck and need some help solving this proof.

    x^log(Y) = y^log(X)

    i am not really sure where to start..

    i thinking about using the log definition first, something like

    Y = log(X)

    X = b^Y

    and then taking log of both sides to get

    log(X) = log(Y)

    though, i am pretty lost from here. I am not even sure if I am doing this right, since I was told to work backwards first, then fowards.

    any help / suggestions would be awesome !
    Last edited by jimbutler83; January 29th 2010 at 11:05 AM.
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  2. #2
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    Quote Originally Posted by jimbutler83 View Post
    Hello, I'm currently stuck and need some help solving this proof.

    x^log_b(Y) = y^log_b(X)

    i am not really sure where to start..

    i thinking about using the log definition first, something like

    ...
    That's exactly the right idea :

    x = b^{\log_b(x)} and y = b^{\log_b(y)}

    Plug in these terms into the given equation:

    x^{\log_b(y)} = y^{\log_b(x)}~\implies~b^{\log_b(x) \cdot \log_b(y)} = b^{\log_b(y) \cdot \log(x)}

    Can you take it from here?
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  3. #3
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    Hello, jimbutler83!

    You're on the right track . . .


    Prove: . X^{\log(Y)} \:=\: Y^{\log(X)} . The base of the logs is not important.

    \text{Let: }\:X^{\log(Y)} \:=\:P .[1]

    Take logs: . \log\left(X^{\log(Y)}\right) \:=\:\log(P) \quad\Rightarrow\quad \log(Y)\!\cdot\!\log(X) \:=\:\log(P)

    We have: . \log(X)\!\cdot\!\log(Y) \:=\:\log(P) \quad\Rightarrow\quad \log\left(Y^{\log(X)}\right) \:=\:\log(P)

    Hence: . Y^{\log(X)} \:=\:P .[2]


    Equate [1] and [2]: . X^{\log(Y)} \;=\;Y^{\log(X)}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, jimbutler83!

    You're on the right track . . .



    \text{Let: }\:X^{\log(Y)} \:=\:P .[1]

    Take logs: . \log\left(X^{\log(Y)}\right) \:=\:\log(P) \quad\Rightarrow\quad \log(Y)\!\cdot\!\log(X) \:=\:\log(P)

    We have: . \log(X)\!\cdot\!\log(Y) \:=\:\log(P) \quad\Rightarrow\quad \log\left(Y^{\log(X)}\right) \:=\:\log(P)

    Hence: . Y^{\log(X)} \:=\:P .[2]


    Equate [1] and [2]: . X^{\log(Y)} \;=\;Y^{\log(X)}

    thanks so much, though i was wondering if it is possible to explain this step a bit more clearly..

    why does

    \log(X)\!\cdot\!\log(Y) \:=\:\log\left(Y^{\log(X)}\right)

    is this some kind of log property ?


    thanks again.
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  5. #5
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    Quote Originally Posted by jimbutler83 View Post
    thanks so much, though i was wondering if it is possible to explain this step a bit more clearly..

    why does

    \log(X)\!\cdot\!\log(Y) \:=\:\log\left(Y^{\log(X)}\right)

    is this some kind of log property ?


    thanks again.
    Yes.

    a^n~\implies~\log(a^n) = n \cdot \log(a)

    The exponent becomes a factor of the log-value.

    If you have a factor at a log-term this factor can be expressed as an exponent. That's what Soroban did.
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