# Asymptotics of the solution of $x^x=n$

I need to find asymptotics of the solution of the equation $$x^x=n$$ while $n\to\infty$. The only thing I understand is this solution grows very slowly. I can't find $x$ explicitly, I think this is impossible. So what is the necessary trick?

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Will $x = o (\log n)$ suffice? – user17762 Jun 28 '12 at 23:50
Take logarithms on either side: $x\log(x)=\log(n)$. From there you should be able to bootstrap. – Steven Stadnicki Jun 28 '12 at 23:52
Relevant: Lambert W-Function – user2468 Jun 28 '12 at 23:56

## 2 Answers

How about $$x = \frac{\operatorname{ln} (n)}{\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)} + \frac{\operatorname{ln} (n) \operatorname{ln} \bigl(\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)\bigr)}{\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)^{2}} - \frac{\operatorname{ln} (n) \operatorname{ln} \bigl(\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)\bigr)}{\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)^{3}} +\dots$$ You can find computaiton of asymptotics for the Lambert W function as Problem 4.2 in my paper Transseries for Beginners

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I didn't knew that my question touches such a deep mathematical theories – user34574 Jun 29 '12 at 7:08

As the other posters already pointed out, the Lambert W function is your friend here. If you take log on both sides then you get $$x \log(x) = \log(n)$$ If you assume that $x > 0$, then you can write $x = e^y$ (for $y = \log x$). Then it becomes $$e^y y = \log n$$ which has $y = W(\log n)$ as a solution, where $W$ is the (principal branch of the) Lambert W function. Thus the solution to your original problem is $x = e^{W(\log n)}$, which grows to $\infty$ as $n \rightarrow \infty$ (although very, very slowly: $e(W(\log(10^6))) \approx 7$).

And you can even simplify this solution since (by definition) $$e^{W(\log n)} W(\log n) = \log n$$ and therefore $$x = e^{W(\log n)} = \frac{\log n}{W(\log n)}.$$

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