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Math Help - Solving Exponential Equations

  1. #1
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    Solving Exponential Equations

    Hi everyone,

    I am stumped on a question that I have been wondering about for the whole day. I know that 2^2+3^2=13.. but I wondered what would happen if the exponents were variables....

    2^x+3^x=13


    Would there be any possible way to solve this equation?

    Thanks ahead.
    Last edited by KelvinScale; October 6th 2009 at 02:55 PM. Reason: Typed exponents incorectly
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  2. #2
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    Yes there is, with the help of the logarithm function.

    log A^B= B log A

    so take the log of both sides to get:  log2^x + log3^x = log13

    Now apply the property log A^B= B log A and you would be able to solve for x.
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  3. #3
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    Hmm.. it doesn't seem to work, because I get 1.4 something if I use that method.
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  4. #4
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    Sorry, my mistake. log only works if you want to solve 2^x = 13.

    Other than trial and error, I don't know any other way. Sorry.
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  5. #5
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    They're different bases, aren't they? In a problem I had tried to take away 4^x-2^x but the person said you can't do it if they're different bases.

    Idk though. It's a Pythagorean triple.
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