After looking at it for a while it seems quite a hard problem. I am not an expert in the area, but I have the impression that it might me an open problem.
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$\begingroup$ You might think, absurdly, that $3^{\log_{10}(3)}$ is a rational of the $m/n$ type. $\endgroup$– SebastianoSep 22, 2019 at 22:40
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$\begingroup$ Where does the problem come from? $\endgroup$– saulspatzSep 22, 2019 at 22:46
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$\begingroup$ @saulspatz just a friend of mine posing the question, it seems a rather natural/well-known type of question. $\endgroup$– Martin OrtizSep 22, 2019 at 22:52
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2 Answers
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This would follow from Schanuel's conjecture. It implies in particular that there are no "non-obvious" algebraic relations between logarithms of algebraic numbers.
Assume that $3^{\log_{10} 3}= r$, with $r$ algebraic. Then we would have $$\frac{\log r}{\log 3}=\frac{\log 3}{\log 10}$$ that is, $$(\log 3)^2 = \log r \log 10$$
Now, $\log r$ would have to be a rational combination of $\log 3$, $\log 10$, so this implies an algebraic relation between $\log 3$ and $\log 10$, not possible.
answered Sep 22, 2019 at 23:25
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According to Alan Baker (A concise introduction to the theory of numbers), Lindemann proved in his proof of the trascendance of $\pi$, several other facts, in particular the trascendance of $\log(\alpha)$ for algebraic $\alpha\ne0,1$.(this theorem says: For algebraic numbers, distinct, $\alpha_1,\alpha_2,\cdots,\alpha_n$ and arbitrary algebraic numbers $\beta_1,\beta_2,\cdots,\beta_n$ distinct of zero, one has $\sum\beta_ie^{\alpha_i}\ne0$).
Taking this into account $\log 3$ is trascendent and to $x=3^{\log(3)}$ does not apply Gelfond-Schneider theorem. As far as I know $x$ can be rational (even integer) or irrational algebraic or trascendent if we want to apply the known theory.
