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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?

itdoesntwork
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  • Elementary application of the lambert w function. – Listing Dec 17 '11 at 00:21
  • You might want to look into the Lambert $W$-function Some manipulation will put your equation into the requisite form. The Lambert $W$-function cannot be expressed in closed form in terms of the usual elementary functions. Some mathematics packages, for example Maple, have it routinely available to compute with. – André Nicolas Dec 17 '11 at 00:23
  • Alright, thanks for the clarification. I thought there might be a way to do without the product log. – itdoesntwork Dec 17 '11 at 00:30

2 Answers2

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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)

AMPerrine
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Listing
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  • Thanks! I was solving that equation because it yields the solution to the infinite power tower problem: Given a number, raise it to the power of itself infinite times. I guess the only good ways are numerical. – itdoesntwork Dec 17 '11 at 00:37
  • I see, that is example 3 in the wikipedia article. It yields the same result :) – Listing Dec 17 '11 at 00:41
  • @itdoesntwork: I presume you haven't seen this... – J. M. ain't a mathematician Dec 17 '11 at 01:55
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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... – J. M. ain't a mathematician Dec 17 '11 at 01:58
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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.

Bill Cook
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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! – itdoesntwork Dec 17 '11 at 00:28