# Algebraic number that exponentiated with algebraic number give $\pi$

I'm not sure if an algebraic number elevated with an algebraic exponent can give rise to a transcendental number.

If that's the case does anybody know a closed form for an algebraic number that exponentiated with an algebraic exponent gives $\pi$?

• Yes: $(\sqrt[e]{\pi})^e$, since $\sqrt[e]{\pi}$ is obviously "rational or irrational", and $e$ is obviously irrational (assuming that by "exponentiated with" you mean "raised to a power of"). – barak manos Jul 15 '16 at 10:50
• Another example is $x^x = \pi$. The unique $x$ solving this equation is clearly irrational (otherwise $\pi$ would be algebraic). So there is certainly an irrational number, which, raised to an irrational power, gives $\pi$. Furthermore, for any positive rational $q\neq 1$ let $z$ be such that $q^z = \pi$. Then, again, $z$ is irrational, so any positive rational number (except $1$, naturally) can be raised to an irrational power to get $\pi$. – zhoraster Jul 15 '16 at 10:59
• To the edited question. Good question. I've never heard of such a number – Yuriy S Jul 15 '16 at 11:12
• I believe this is an open question. But you're essentially asking if this expression: $\frac{\ln \pi}{W(\ln \pi)}$ is algebraic (where $W(x)$ is the Lambert W function). This is very likely unknown (though it is definitely irrational, as pointed out by @zhoraster). – Deepak Jul 15 '16 at 11:29