[manmanic]

2=x^x^x^x... cannot be solved, since x^x^x^x... will only converge to 0 (if x=0) or 1 (0<x<=1) or -1 (x=-1), ignoring complex numbers. So the original equation is false.

[BoppreH]

Wikipedia says it (roughly) converges if e^(−e) ≤ x ≤ e^(1/e)

http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite...

[RiderOfGiraffes]

As a simple counter-example to your claim, the sequence:

  x, x^x, x^x^x, x^x^x^x, ...

when x = sqrt(2) is strictly increasing and bounded above, and therefore converges. It's not hard to show that it's bounded above by 2, because x^x^2 > x^x^x, and x^x^2 = x^2 = 2. Repeat for any length sequence of exponentiation.

[fxj]

maybe you meant 4=x^x^x^x^x^x... cannot be solved because there exists no x that solves the equation y=x^x^x^x^x... for a maximum value of y which is a positive real solution of the equation y = x^y. Thus, x = y^(1/y). The limit defining the infinite tetration of x fails to converge for x > e^(1/e) because the maximum of y^(1/y) is e^(1/e). The maximum value is smaller than 4 and thus no solution exists.