# Thread: Solving ln(x) = kx

1. ## 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?

2. Originally Posted by dh273
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.

3. 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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