How solve for x in an Infinite exponent How would one solve for x in the following equation:

$x^{x^{x^{x^{\cdots}}}} = 4$

The exponent continues forever... So what is the value of x?

Thank you for helping
An interesting dynamical system is hidden behind the curtain. In the socalled power tower previously mentioned one usually looks at $0&lt;x&lt;1$ for which the map $f_x(y)=x^y$ is decreasing and there is a well-defined meaning to the tower-limit which is simply the unique fixed point of the map. There is a catch to the problem, however, when $x&gt;1$. In that case there are two fixed points for $1&lt;x&lt;1.444...$ In the case mentioned, starting with $y=x$, and iterating $y\mapsto x^y$ you will end up at the fixed point $y=2$ and not 4 as expected! On the other hand $y=4$ is indeed a fixed point, though unstable.  Try on a pocket calculator or similar (not maple, though) to iterate $y\mapsto (\sqrt{2})^y$ starting with $y=4$. Takes about 50 iterations before you see that the fixed point is indeed unstable.
The difficulty is perhaps: how do you define $x^{x^{x\ldots}}$?  

We could define it as the limit of successive towers, defining the sequence $y_{n}=x^{y_{n-1}}$ and $y_{1}=x$. Then, assuming this sequence has a limit (!), this limit satisfies $x^{y}=y$, by continuity of exponentiation.  

If we insist, say, that $x$ is positive, then the equation $y=4$ is equivalent to the equation $x^{y}=x^{4}$, i.e. $x^{4}=4$. This has the positive solution $x=\sqrt{2}$.