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Math Help - Exponential function disguised as a quadratic

  1. #1
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    Exponential function disguised as a quadratic

    5^{2x} -130^x + 25 = 0

    This is what I tried:

    5(1^{2x}-26^x + 5) = 0

    5(1^{2x}-26 * 1^x + 5) = 0

    1^x = y

    5(y^2-26y+5) = 0

    Solving the above quadratic doesn't give me the right answers, unless I solved it wrong, but I don't think I did. What did I do wrong?
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  2. #2
    Newbie Mirado's Avatar
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    a little help

    Fatal mistake, the way you are factoring the exponential pieces. You can not factor the 5 out like that, or the 1 like that.
    Last edited by Mirado; April 20th 2009 at 10:35 PM. Reason: wrong heading
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by Phire View Post
    5^{2x} -130^x + 25 = 0

    [snip]
    Are you sure this is the correct equation? I'll bet it's meant to be 5^{2x} - 26 \cdot 5^x + 25 = 0 but you made an incorrect 'simplification' to this before posting.

    If so then you're meant to factorise it as (5^x - 25)(5^x - 1) = 0. Now solve for x.
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  4. #4
    Super Member Gamma's Avatar
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    I came to the same conclusion today while at school, this problem has been bugging me for 24 hours. Me and 4 other grad students worked on it over lunch to no avail, no way the problem posted as is should be assigned in a pre calculus class, maybe a numerical methods course at best.

    If it is as Mr. Fantastic suggested (and I hope it is) simply substitute y=5^x and solve the quadratic y^2 - 26y + 25 = 0to see how he got his factoring, then you are basically done.
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