The solution of $x^x=2$ rational/algebraic irrational/transcendental?

Question

What does the unique real number $x$ such that $x^x=2$ equal to? Is the value rational, algebraic irrational or transcendental? What about $x^x=3$? Or $x^x=e$? $x^x=π$?

Answer 1

The Gelfond-Schneider Theorem tells you that $x$ cannot be an algebraic irrational in the case of $2$ and $3$, but says nothing in the case of $e$ or $\pi$. For elementary reasons $x$ could not be rational, since $(\frac{a}{b})^a=2^b$ is impossible as the right-hand side is an integer, and the left-hand side is not. There may not be much else known.

Answer 2

One may find solutions in terms of the Lambert W function:

$$x^x=a\\\ln(x)e^{\ln(x)}=\ln(a)\\\ln(x)=W(\ln(a))\\x=e^{W(\ln(a))}=\frac{\ln(a)}{W(\ln(a))}$$

In the case that we are solving $x^x=e$, we find that

$$x=e^{W(1)}=e^\Omega=\frac1\Omega$$

where $\Omega$ is the Omega constant, which is known to be transcendental, thus $x$ is transcendental. It follows directly from the Lindemann-Weierstrass theorem that if $\frac1\Omega$ were algebraic, then $e^\Omega$ would be transcendental, a contradicition, so $\Omega$ must be transcendental.

Similar arguments may be constructed to show that if $a=e^t$ for algebraic $t$, then $x$ is transcendental.