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Math Help - ln(x)=e^-x find x

  1. #1
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    ln(x)=e^-x find x

    ln(x)=e^(-x)

    find x...
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  2. #2
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    Quote Originally Posted by ice_syncer View Post
    ln(x)=e^(-x)

    find x...
    Hi ice_syncer,

    you could use Newton's Method for finding roots.

    f(x)=e^{-x}-lnx

    The solution to lnx=e^{-x}

    is the root of f(x)=0

    Take an initial shot at the solution as x=1.6

    Then

    x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=1.6-\frac{e^{-1.6}-ln(1.6)}{-e^{-1.6}-\frac{1}{1.6}}=1.275767

    A few more iterations would close in on the root, approximately 1.31
    Attached Thumbnails Attached Thumbnails ln(x)=e^-x    find x-nm.jpg  
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  3. #3
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    Quote Originally Posted by ice_syncer View Post
    ln(x)=e^(-x)

    find x...
    Please post the whole question (in particular the background to it that probably includes something like "Find the approximate solution of ....")
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    how did you think of the initial shot ie x = 1.6
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    Quote Originally Posted by ice_syncer View Post
    how did you think of the initial shot ie x = 1.6
    e^{-x} is a decreasing function, while lnx is increasing.

    e^{-1}=0.368

    ln1=0

    e^{-2}=0.135

    ln2=0.693

    Hence these cross between x=1 and x=2.

    You could choose any value of x in the vicinity as a starting point, say x=1.5 or so. Choosing x=1 or x=2 would also suffice.

    The attachment shows Newton's method homing in on the solution.
    Attached Thumbnails Attached Thumbnails ln(x)=e^-x    find x-nm2.jpg  
    Last edited by Archie Meade; June 9th 2010 at 07:07 AM.
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    .......ok... I'll read more about it, thanks!
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