Transcendence of $\pi+\log\alpha$ and $e^{\alpha\pi+\beta}$

Question

In the book "Transcendental Number Theory" by Alan Baker, he proves a few corollaries of Baker's theorem. I've attached this page below.

After, he claims that special cases of these corollaries give the transcendence of $\pi+\log\alpha$ for $\alpha\neq 0$ and $e^{\alpha\pi+\beta}$ for $\beta\neq 0$. I'm unable to see how one establishes this.

For $\pi+\log\alpha$ I've tried assuming it is algebraic and using Theorem 2.3 to conclude that $e^{\pi+\log\alpha}=\alpha e^{\pi}$ is transcendental. But this isn't very useful.

I'm not sure about $e^{\alpha\pi+\beta}$.

This seems really trivial and I'm just missing something. Any help on seeing this would be appreciated.

Answer 1

Since $e^{i\pi}=-1$ it can be seen that $\pi=\frac{\ln(-1)}{i}$ and since $i$ and $-1$ are algebraic, $\pi+\ln(\alpha)$ for algebraic $\alpha$ is transcenental by Theorem 2.2.

Using $\pi=\frac{\ln(-1)}{i}$ again, $e^{\alpha\pi+\beta}=e^\beta(-1)^\frac{\alpha}{i}$ which is transcendental by Theorem 2.3.