What is the best asymptotic approximation of the inverse $x=g(y)$ of $y = x^x$ for large $x$? [Clearly, if $x>e$, then $f(x) > e^x$ implies $g(x) < \log x$.]
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2$\begingroup$ $g(y) = e^{W(\log y)}$ where $W(\cdot)$ is the Lambert W-function. Asymptotics of the Lambert W function is probably well studied. The Wikipedia article on the Lambert W function would be a good place to start. $\endgroup$– DineshCommented Apr 4, 2010 at 3:19
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3$\begingroup$ Mathworld(mathworld.wolfram.com/LambertW-Function.html) gives an asymptotic expression which if restricted to the first two terms gives $x \approx \frac{\log y}{\log \log y}$ as noted in gowers' answer. You can add more terms if need be to reach your desired level of accuracy. $\endgroup$– DineshCommented Apr 4, 2010 at 3:33
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I don't know how accurate you want to be, but a quick and dirty approximate inversion of $x\log x$ is $x/\log x$. So if $y=x^x$ then $\log y\approx x\log x$, so $x\approx\log y/\log\log y$. But perhaps you want something better than this.
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$\begingroup$ sorry, why is approx. inversion of xlogx is x/logx ? $\endgroup$– Chan KimCommented Jun 9, 2016 at 2:44