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If $x^x=n$, how does $x$ grow asymptotically in terms of $n$?

We know that it grows slower than $\log n$, because $\log n^{\log n}>e^{\log n}=n$. But it grows faster than $\log \log n$, because $\log\log n^{\log\log n}<n$.

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Simple manipulations show that $$x=\frac {\log n}{W(\log n)} \sim \frac {\log n}{\log\log n} $$ where W is Lambert W function.

Edit: $x\log x=\log n\Rightarrow \log x = W(\log n)\Rightarrow x=e^{W(\log n)}=\frac {\log n}{W(\log n)}\sim\frac {\log n}{\log\log n-\log\log\log n}$.

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  • $\begingroup$ Could you please show the manipulations, and why $W(\log n)\sim \log\log n$? Also, I think you meant $\log n$ instead of $\log x$. $\endgroup$ – Alexi Oct 13 '15 at 4:33
  • $\begingroup$ $x\log x=\log n\Rightarrow \log x = W(\log n)\Rightarrow x=e^{W(\log n)}=\frac {\log n}{W(\log n)}$. Also $W(x)\sim \log x$ $\endgroup$ – A.S. Oct 13 '15 at 4:39

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