Given that

  1. $x^x = y$; and
  2. given some value for $y$

is there a way to expressly solve that equation for $x$?

  • 5
    $\begingroup$ Take logs, set $x = e^t$ and apply this: math.stackexchange.com/questions/10261/inverse-of-y-xex $\endgroup$ – Aryabhata Jul 28 '11 at 7:03
  • $\begingroup$ For $e^{-1/e} \lt y \le 1$ there will be two non-negative solutions for $x$ and for $y \lt e^{-1/e}$ there will be none $\endgroup$ – Henry Sep 5 '16 at 10:18
  • $\begingroup$ It is also known as the super-square root of $y$. $\endgroup$ – Simply Beautiful Art Dec 31 '16 at 1:00
  • $\begingroup$ 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 $x=e^{W(\ln y)}=\frac{\ln y}{W(\ln y)}$. $\endgroup$ – Ahmed S. Attaalla Jan 9 '17 at 5:56

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:

$$x=\frac{\ln(y)}{\ln{\ln y} - \ln\ln{\ln y} + o(1)} = \Theta\left( \frac{\ln y}{\ln \ln y}\right)$$

  • $\begingroup$ Wow - thanks - I really didn't think it was possible. $\endgroup$ – Josh Jul 28 '11 at 22:23
  • $\begingroup$ @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. $\endgroup$ – Yves Daoust Jul 9 '18 at 14:04

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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  • 4
    $\begingroup$ If you're just going to post a link to wolframalpha, you could at least make sure that it works... $\endgroup$ – t.b. Dec 31 '11 at 16:20
  • 1
    $\begingroup$ If a natural number y is entered in "solve x^x=y" WA will give the numeric answer. Strangely it doesn't work for all real y. However, I downvoted since the link is no help in understanding how the solution was reached. $\endgroup$ – David Marquis Dec 31 '11 at 17:22
  • 2
    $\begingroup$ I guess part of GSBabil's point is that Josh should have tried that first, before posting here. $\endgroup$ – GEdgar Dec 31 '11 at 18:33

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