# 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.

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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. – Sangchul Lee 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 $e^{\frac{ln(2)}{2}}=e^{ln(\sqrt{2})}=\sqrt{2}$. – T. Eskin 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