Why does $$\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\cdots}}}}}}}}}=\log_x{e}=\frac{1}{\ln{x}}$$ There only seems to be a relation when using square roots, but not for cubed roots or anything else. Why does this equation work and why does it only work for square roots?
(The $e$ is not significant, by the way. You could give the exponential function a different base, $a$, and say the equation equals $log_x{a}$).
On
Elaborating on what Jack said, assume we have an $n$th root instead of a square root:
$$y = \sqrt[n]{\log_x{\exp{\sqrt[n]{\log_x{\exp{\sqrt[n]{\log_x{\exp{\cdots}}}}}}}}}$$
Then
$$y = \sqrt[n]{\log_x{\exp\left(y\right)}}$$
$$y = \sqrt[n]{y\log_x{e}}$$
$$y^n = y\log_x{e}$$
$$y^{n-1} = \log_xe$$
Obviously, with $n = 2$, $n-1 = 1$, meaning $y$ itself equals $\log_xe$.
This can be expanded upon though.
$$y=\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\sqrt{\log_x{\exp{\cdots}}}}}}}}}\implies y=\sqrt{\log_x\exp(y)}=\sqrt{y\log_xe}\\ \therefore y=\log_xe$$