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Given that

  1. $x^x = y$; and
  2. given some value for $y$

is there a way to expressly solve that equation for $x$?

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Take logs, set $x = e^t$ and apply this: – Aryabhata Jul 28 '11 at 7:03
up vote 17 down vote accepted

As Aryabhata mentions this is another application for the Lambert W function. The solution to your problem is presented in the wikipedia article. Using elementary substitutions you have

$$x=\frac{\ln(y)}{W(\ln y)}$$

If you are interested in the asymptotic growth of $x$ relative to $y$, note that for every $z$: $W(z) = \ln{z} - \ln\ln{z} + o(1)$. Hence:

$$x=\frac{\ln(y)}{\ln{\ln y} - \ln\ln{\ln y} + o(1)} = \Theta\left( \frac{\ln y}{\ln \ln y}\right)$$

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Wow - thanks - I really didn't think it was possible. – Josh Jul 28 '11 at 22:23

You should try WolframAlpha for similar problems. WolframAlpha would solve y=x^x for y=5 as shown here (using Lamber W Function as suggested before).

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If you're just going to post a link to wolframalpha, you could at least make sure that it works... – t.b. Dec 31 '11 at 16:20
If a natural number y is entered in "solve x^x=y" WA will give the numeric answer. Strangely it doesn't work for all real y. However, I downvoted since the link is no help in understanding how the solution was reached. – David Marquis Dec 31 '11 at 17:22
I guess part of GSBabil's point is that Josh should have tried that first, before posting here. – GEdgar Dec 31 '11 at 18:33

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