# How to solve infinite repeating exponents

How do you approach a problem like (solve for $x$):

$$x^{x^{x^{x^{...}}}}=2$$

Also, I have no idea what to tag this as.

Thanks for any help.

-
Assume first that the identity holds. The you will have $x^2=2$. Since $x$ should be positive, we have $x=\sqrt{2}$. Now it remains to prove that this value is really a solution. You may consider a sequence $a_{n+1} = \left(\sqrt{2}\right)^{a_n}$ to give a rigorous proof. –  sos440 Jul 4 '12 at 6:25
Thanks for this comment. I overcomplicated this problem. –  Matt Jul 4 '12 at 6:52

I'm just going to give you a HUGE hint. and you'll get it right way. Let $f(x)$ be the left hand expression. Clearly, we have that the left hand side is equal to $x^{f(x)}$. Now, see what you can do with it.

-
Note, though, that the argument that you have in mind only shows what the only possible solution is. The argument fails if the righthand side is $4$, say, as there is then no solution. The lefthand side converges iff $e^{-e}\le x\le e^{1/e}$. –  Brian M. Scott Jul 4 '12 at 6:32
$x^{f(x)} = 2, f(x)ln(x)=ln(2), f(x) = ln(2)/ln(x)$, since $f(x) = 2, ln(2)/ln(x) = 2$, solving for x gives $e^{ln(2)/2}$? –  Matt Jul 4 '12 at 6:38
@Matt. Dito. And even further, you may notice that $e^{\frac{ln(2)}{2}}=e^{ln(\sqrt{2})}=\sqrt{2}$. –  Thomas E. Jul 4 '12 at 6:42
I see. Thanks for the hint/help! –  Matt Jul 4 '12 at 6:42
Correcting my earlier comment: It does not fail when the righthand side is $4$. The restriction on $x$, however, is correct. –  Brian M. Scott Jul 4 '12 at 6:57