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Math Help - A bit of an exponential pickle.

  1. #1
    No one in Particular VonNemo19's Avatar
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    A bit of an exponential pickle.

    x^2+e^{-x}=8

    Is there an analytic solution? If not, why so?

    e^xx^2+1=8e^x\Rightarrow{e}^x=\frac{-1}{x^2-8} ?

    All this tells me is that |x|>\sqrt8
    Last edited by VonNemo19; July 29th 2009 at 09:02 PM.
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post
    x^2+e^{-x}=8

    Is there an analytic solution? If not, why so?

    e^xx^2+1=8e^x\Rightarrow{e}^x=\frac{-1}{x^2-8} ?

    All this tells me is that |x|>\sqrt8
    I didn't check your calculations but from the last equation you know that -\sqrt{8} < x < \sqrt{8}:

    e^x is positive
    Therefore the denominator of the fraction must be negative

    EDIT:

    Probably you have noticed that your last equation can't be used to solve it for x because you have the variable on both sides of the equation.

    I "solved" (using Newton's method) the equation:

    x^2+e^{-x}-8=0

    and I got 2 real approximate results: x_1 \approx 2.817847...~\vee~x_2\approx -1.65826...
    Last edited by earboth; July 29th 2009 at 09:42 PM.
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