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Thread: 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.

    $\displaystyle x^2+e^{-x}=8$

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

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

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

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

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

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

    $\displaystyle e^x$ 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:

    $\displaystyle x^2+e^{-x}-8=0$

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