# Is $3^{\log_{10}(3)}$ irrational?

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.

• You might think, absurdly, that $3^{\log_{10}(3)}$ is a rational of the $m/n$ type. Sep 22, 2019 at 22:40
• Where does the problem come from? Sep 22, 2019 at 22:46
• @saulspatz just a friend of mine posing the question, it seems a rather natural/well-known type of question. Sep 22, 2019 at 22:52

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.
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.