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Here is my question. Are there $\alpha$ and $\beta$ transcendental numbers such that both $\alpha^{\beta}$ and $\beta^{\alpha}$ are algebraic?

There isn't anything specific about roots of the question. (I was just thinking over).

Of course $e^{\ln2} = 2$, but what can we say $(\ln 2) ^{e}$ ?

If the answer is yes, what can we say about number of such pairs. (Are the set of these pairs finite ?)

After helpful comments let me make this amendment. $\alpha > \beta$.

kolobokish
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    Well $\alpha=\beta$ might be a good place to start. Is the real solution to $\alpha^{\alpha}=2$ transcendental? It is definitely irrational, but I can’t quite prove it is irrational. – Thomas Andrews Sep 25 '21 at 17:37
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    @ThomasAndrews If $\alpha$ were irrational algebraic, $\alpha^{\alpha}$ would be transcendental (Gelfond Schneider , $\alpha \ne 0$ and $\alpha \ne 1$ is obvious). Hence $\alpha$ must in fact be transcendental. – Peter Sep 25 '21 at 17:42
  • @ThomasAndrews That a good question, if we can find some algebraic number a for which $\alpha^{\alpha}=a$ has transcendental solution, we are done. (Thank you, I'll think over it). One more point, even if we find one, it seems to me we can't say that there are other such numbers. – kolobokish Sep 25 '21 at 17:43
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    Distinct numbers would be nice. – Peter Sep 25 '21 at 17:44
  • @Peter Thank you. Initially in my mind I was bearing exactly distinct number. )). But is this question even interesting to mathematical society? I'm not sure if it has any importance. Is even exponentation to transcendence degree important topic? – kolobokish Sep 25 '21 at 17:46
  • Related: https://math.stackexchange.com/questions/3078472/is-there-any-known-transcendental-b-such-that-bb-is-also-transcendental?rq=1 – Mason Oct 22 '21 at 19:17

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