I am looking at the rationality of $\log_ba$ when at least $a$ or $b$ is irrational such that $a,b>1$ and neither of $a$ or $b$ is of the form $\left(\frac{e}{f}\right)^{(g/d)}$, where $d,e,f,\text{ and }g\in\mathbb{N}$.
I know how to check irrationality when both of $a$ and $b$ are rational, one of the beautiful answer by Gone is here.
I am done with the case when exactly one of them is irrational.
Here my version of proof:
W.L.O.G, Assume $a$ is irrational.
Suppose that $$\log_ba=p/q\enspace\text{ where } p,q\in\mathbb{N}\\\implies a^q=b^p$$ but $a^q$ is irrational and $b^p$ is rational. Hence, contradiction.
Also, if $a=b^c$, then it depends on $c$.
Finally, when $a$ and $b$ are irrational and $a\neq b^c$.
I have no idea about it. Can anyone help me with this? Also, is my above proof correct? If incorrect, please do provide the correct one. Thanks.
If $a \not = b^c$ for $c \in \mathbb{Q}$, then $\log_b a = c \not \in \mathbb{Q}$. Now if we reduce the assumption to $c \in \mathbb{Z}$:
Let $a=\sqrt{\pi}$ and $b=\pi$. Then since neither $a,b$ are algebraic nor rational, they fit your assumption that neither equal $\left(\frac{e}{f}\right)^{(g/d)}$. $$\log_{b}a=\log_{\pi}\sqrt{\pi}=\frac{1}{2} \in \mathbb{Q}$$