Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


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

Thanks for any help.

share|cite|improve this question
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

1 Answer 1

up vote 8 down vote accepted

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.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.