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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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    Relevant: [Lambert W-Function](http://en.wikipedia.org/wiki/Lambert_W_function)2012-06-28

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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 theories2012-06-29
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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)}. $