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Math Help - Logarithmic equation

  1. #1
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    Question Logarithmic equation

    Please help! Many thanks in advance! I can't find any analytical solution to the following:


    Solve \log_5 (x-4) = \log_7 x for x. Round the answer from your calculator to 4 decimal places.

    This is a "Show your work" question. I don't understand what work can be shown here except for the final answer: x=11.5842
    Last edited by Ivan; September 17th 2007 at 10:53 AM.
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  2. #2
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    Quote Originally Posted by Ivan View Post
    This is a "Show your work" question.
    Well, then you had better start showing! What have you tried? Do you have the change of base formula?

    For appropriate a, b, and c

    log_{b}(a) = \frac{log_{c}(a)}{log_{c}(b)}

    That should make you feel like you are doing something. Too bad it will not lead anywhere.

    Since you must use iterative techniques to solve this, you should state what method you are using and then implement it carefully.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Can someone please do this question? I've tried some crazy stuff to try and work it out, all failed except one where I ended up using the Newton-Raphson method, which I think is too much work. Is there some easier way I am overlooking?
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  4. #4
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    Quote Originally Posted by Ivan View Post
    Solve \log_5 (x-4) = \log_7 x for x. Round the answer from your calculator to 4 decimal places.

    This is a "Show your work" question. I don't understand what work can be shown here except for the final answer: x=11.5842
    I don't find an analytical solution yet (actually, I already quit to this), but try graphing each side, you'll find the solution.
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  5. #5
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    Say,
    \frac{\log (x-4)}{A} = \frac{\log x}{B}
    Where A,B>0.

    10^{\log(x-4)/A} = 10^{\log x/B}

    (x-4)^{1/A} = x^{1/B}

    Thus,
    (x-4)^B = x^A

    There is no general analytic approach in general if A\not = B and they are not integers.
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