We know existence of Transcendental raised to Algebraic Irrational equals rational, but what about opposite?

Introduction:

If we take $$a=2^\sqrt[3]{2}$$ which is transcendental by Gelfond-Schneider Theorem, and $$b=\sqrt[3]{4}$$ which is algebraic irrational because it is root of monic-irreducible polynomial over $$\mathbb{Q}$$ of degree $$3$$ which is $$x^{3}-4$$.(The polynomial is irreducible by the theorem: If polynomial in $$F[x]$$ where $$F$$ is field has degree $$2$$ or $$3$$ and doesn't have roots in $$F$$, then it is irreducible over $$F$$ ).

So, there exist transcendental $$a$$ and algebraic irrational $$b$$ such that $$a^{b}$$ is rational.

My question: Can the opposite happen, i.e, there exist algebraic irrational $$a$$ and transcendental b, such that $$a^{b}$$ is rational?

Yes. Let $$a=\sqrt{2}$$. By the intermediate value theorem, there is some $$0, for which $$\sqrt{2}^x$$ is rational. $$x$$ can not be rational, and, by Gelfond-Schneider, it can not be algebraic either so it must be transcendental. (As in the comments you can solve for $$x$$ if you want, but you don’t need to.)
• The following is just discussion, so irrational to irrational is rational in the following cases: 1) Transcendental to algebraic irrational, one of the examples is the one I mentioned in the body of my question. 2)Algebraic irrational to transcendental, one of the examples is in your answer, one specific example is the solution to $(\sqrt{2})^{x}=\frac{7}{5}$, $x=2*\frac{ln(\frac{7}{5})}{ln(2)}$. 3) Transcendental to Transcendental, one of the examples is $e^{ln(2)}$. Note, Algebraic irrational to Algebraic irrational is not included by Gelfond-Schneider Theorem. Commented Jun 19 at 19:41
• You have not worked out why $x$ cannot be rational , and it is not at all obvious. Commented Jun 19 at 20:04
• @Peter Consider a rational in reduced form, then $\sqrt{2}^{p/q}$ being rational must mean that $2^p$ is a $2q$-th perfect power, so $q$ divides $p$, meaning that $p$ is an integer…. Commented Jun 19 at 20:11