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Math Help - Solving ln(x) = kx

  1. #1
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    Solving ln(x) = kx

    Hi. I'm trying to find an analytical solution to ln(x) = kx, where k is a constant. I can solve it with numerical methods (Newton-Raphson, etc), but was trying to find an analytical solution. Does anyone have any ideas?
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  2. #2
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    Quote Originally Posted by dh273 View Post
    Hi. I'm trying to find an analytical solution to ln(x) = kx, where k is a constant. I can solve it with numerical methods (Newton-Raphson, etc), but was trying to find an analytical solution. Does anyone have any ideas?
    You can solve \ln x = kx \Rightarrow x = e^{kx} using the Lambert W-function (http://en.wikipedia.org/wiki/Lambert_W_function. You'll find other links if you search MHF):

    x = e^{kx} \Rightarrow x e^{-kx} = 1 \Rightarrow -kx e^{-kx} = -k \Rightarrow -kx = W(-k) \Rightarrow x = -\frac{W(-k)}{k}.


    NB: Obviously there are restrictions on k for a real solution to exist.
    Last edited by mr fantastic; November 10th 2008 at 01:48 AM.
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  3. #3
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    OK, I took a look at this, and only get one of the three solutions through this method. The exact equation I was trying to solve was 2^x = x^2, which equates to ln(x) / x = ln(2)/2, and rearranging for the W function, you get -x ln(2) / 2 = W[-ln(2) / 2] = -ln(2), so x=2. However, there are also solutions of x=4 and x=-0.7666.. that are not given by this method. Do you know why this is the case?
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