What are all the values of $a$ and $b$ such that $a, b\notin\mathbb{Q}$ and $a^b\in\mathbb{Q}$? For instance, the classic example is $\left(\sqrt2^\sqrt2\right)^\sqrt2=2$. We also have $a,b\notin\mathbb{Q}$ and $b=\log_ak$, where $k\in\mathbb{Q}$, (making sure that $b$ is irrational), then we can see $a^b=k$. What other ways are there to find these $a$'s and $b$'s?

I'm not asking for just one singular answer, but sort of where to go from here to come up with irrational numbers that satisfy everything.

Edit: As Vadim pointed out in the comments, anything like $\left(\sqrt[m]n^\sqrt[k]{m^a}\right)^\sqrt[k]{m^b}$, where $a+b=k$ will work.

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    $\begingroup$ en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem $\endgroup$ – vadim123 Nov 25 '16 at 0:53
  • $\begingroup$ Unless I'm wrong, all the Gellfond-Scneider Theorem does is remove all the algebraic irrational numbers from the set of possible answers. It says nothing about any transcendental numbers, and it even gives the $\sqrt2$ example in the article. $\endgroup$ – AlgorithmsX Nov 25 '16 at 1:02
  • $\begingroup$ Correct; GS is not an answer, but does give useful info. $\endgroup$ – vadim123 Nov 25 '16 at 1:03
  • $\begingroup$ Why the downvotes? Am I missing something? $\endgroup$ – AlgorithmsX Nov 25 '16 at 1:05
  • $\begingroup$ For example, $(\sqrt{n}^\sqrt{2})^\sqrt{2}=n\in\mathbb{Q}$, and by GS, $\sqrt{n}^\sqrt{2}$ is transcendental (and hence irrational). $\endgroup$ – vadim123 Nov 25 '16 at 1:05

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