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I've tried everything: Taking $\ln$ of both sides, raising to the power $x$, nothing seems to work.

Is there a way to solve this or am I going to have to use numerical methods instead?

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    Alright, thanks for the clarification. I thought there might be a way to do without the product log.2011-12-17

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Do the substitiution $x=e^t$

Then you have

$(e^t)^{e^{-t}}=y \iff \ln((e^t)^{e^{-t}})=\ln(y) \iff e^{-t} t=\ln(y)$

Set $t=-z$ then

$z\cdot e^z = -\ln(y)$

Now per definition of the Lambert W Function

$z=W(-\ln(y)) \iff t = -W(-\ln(y))$

Therefore your solution is

$x=e^t=e^{-W(-\ln(y))}=\frac{-W(-\ln(y)}{\ln(y)}$

(the last equality comes from the wikipedia article)

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    Listing, demonstrating the last equivalence is easy. You know that $W(t)\exp(W(t))=t$, so $\exp(W(t))=t/W(t)$. Reciprocate both sides and there you go. Having said that, the form that contains the exponential is a bit more useful, as $W(t)/t$ has an unnecessary removable singularity...2011-12-17
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Natural log: $\ln(x^{1/x})=\ln(y)$ so that $\frac{\ln(x)}{x}=\ln(y)$. That's as far as you can get in terms of elementary functions. To finish solving you need something like the "product log" function as mentioned by Wolfram Alpha as it solves the equation.

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    Yeah I know about the product log function. I knew it would go into that at some point, so I asked here for anything else :) Thanks!2011-12-17