How solve for x in an Infinite exponent

Question

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

Answer 1

Since $x^{x^{x^{x^{...}}}} = 4$ , we can replace the exponent $(x^{x^{x^{...}}})$ with 4 (as it shows it is equal to 4)

\begin{align} x^{x^{x^{x^{...}}}} & = 4\\ x^4 & = 4\\ x & = \sqrt[4]{4} \\ & = \sqrt{2} \end{align}

As suggested by @Batman, this link may be useful: http://mathworld.wolfram.com/PowerTower.html :)

Answer 2

This is equivalent to $x^4=4$, so $x=4^{1/4}$

Answer 3

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 $x=\sqrt{2}$.

Answer 4

An interesting dynamical system is hidden behind the curtain. In the socalled power tower previously mentioned one usually looks at $0<x<1$ 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 $x>1$. In that case there are two fixed points for $1<x<1.444...$ 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.