How do you approach a problem like (solve for $x$):


Also, I have no idea what to tag this as.

Thanks for any help.

  • 5
    $\begingroup$ Assume first that the identity holds. The you will have $x^2=2$. Since $x$ should be positive, we have $x=\sqrt{2}$. Now it remains to prove that this value is really a solution. You may consider a sequence $a_{n+1} = \left(\sqrt{2}\right)^{a_n}$ to give a rigorous proof. $\endgroup$ Jul 4, 2012 at 6:25
  • $\begingroup$ Thanks for this comment. I overcomplicated this problem. $\endgroup$
    – Matt
    Jul 4, 2012 at 6:52
  • 2
    $\begingroup$ Possible duplicate of Convergence of tetration sequence. $\endgroup$
    – Xander Henderson
    Jul 2, 2018 at 19:20
  • $\begingroup$ @SangchulLee Why X should be positive? Am I missing something? $\endgroup$
    – kitta
    Jan 4, 2019 at 8:57

1 Answer 1


I'm just going to give you a HUGE hint. and you'll get it right way. Let $f(x)$ be the left hand expression. Clearly, we have that the left hand side is equal to $x^{f(x)}$. Now, see what you can do with it.

  • 3
    $\begingroup$ Note, though, that the argument that you have in mind only shows what the only possible solution is. The argument fails if the righthand side is $4$, say, as there is then no solution. The lefthand side converges iff $e^{-e}\le x\le e^{1/e}$. $\endgroup$ Jul 4, 2012 at 6:32
  • 1
    $\begingroup$ $x^{f(x)} = 2, f(x)ln(x)=ln(2), f(x) = ln(2)/ln(x)$, since $f(x) = 2, ln(2)/ln(x) = 2$, solving for x gives $e^{ln(2)/2}$? $\endgroup$
    – Matt
    Jul 4, 2012 at 6:38
  • 2
    $\begingroup$ @Matt. Dito. And even further, you may notice that $e^{\frac{ln(2)}{2}}=e^{ln(\sqrt{2})}=\sqrt{2}$. $\endgroup$
    – T. Eskin
    Jul 4, 2012 at 6:42
  • $\begingroup$ I see. Thanks for the hint/help! $\endgroup$
    – Matt
    Jul 4, 2012 at 6:42
  • $\begingroup$ Correcting my earlier comment: It does not fail when the righthand side is $4$. The restriction on $x$, however, is correct. $\endgroup$ Jul 4, 2012 at 6:57

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