
  Find $x$ in
  $$ \Large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$$


A trick to solve this is to see that
$$\large
2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}
\quad\implies\quad
2 = x^{\Big(x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}\Big)} = x^2
\quad\implies\quad
x = \pm \sqrt{2}
$$

Are these solutions correct? If not, why? If yes, are there other solutions?



PS: An extension of this discussion can be found in What we can say about $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$?

Might as well...

The power tower $x^{x^\ldots}$ is equivalent to the function $\exp(-W(-\log\,x))$, where $W(z)$ is the Lambert function, in the range $e^{-e}\leq x\leq e^{1/e}$ (as Norbert mentions in the comments; see also equation 13 in the MathWorld entry linked to). $\exp(-W(-\log\,x))$ can be inverted, like so:

$$\begin{align*}
y&=\exp(-W(-\log\,x))\\
-\log\,y&=W(-\log\,x)\\
(\log\,y)\exp(-\log\,y)&=\log\,x\\
\frac{\log\,y}{y}&=\log\,x\\
x&=\exp\left(\frac{\log\,y}{y}\right)\\
x&=\exp\left(\log\,y^{1/y}\right)=y^{1/y}
\end{align*}$$

If $y=2$, then $x=\sqrt2$.



Knoebel's paper establishes the interval of convergence $[\exp(-e),\exp(1/e)]$ for the power tower function, in the case of positive $z$. The paper notes that a full characterization of the region of convergence of $z^{z^\cdots}$ for complex $z$ remains to be done, but Thron, Shell (of Shellsort fame) and others have given partial results. See also this paper by Anderson for another discussion on the convergence of the power tower, this article by Cho and Park, where they discuss the inverses of the function $z^{1/z}$, and this article by Sondow and Marques.
