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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 $x$?

  • 0
    Assume $y$ is positive we have $x^x=y$ so $x\ln x=\ln y$ so $\ln x e^{\ln x}=\ln y$ so $\ln x=W(\ln y)$ so $x=e^{W(\ln y)}=\frac{\ln y}{W(\ln y)}$.2017-01-09

2 Answers 2

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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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    @Josh: in fact this might as well be solved by defining a new ad-hoc function $Z(x)$ such that $Z(x)^{Z(x)}=x$, so that the solution of the equation is $x=Z(y)$; Lambert is defined by $W(x)e^{W(x)}=x$, there is little magic. The main advantage of reducing to the special Lambert function is that it has already been studied.2018-07-09
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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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    I guess part of GSBabil's point is that Josh should have tried that first, before posting here.2011-12-31