- Starting with the result of the Gelfond–Schneider theorem: $$ a^b = c$$ where a and b are complex algebraic numbers with $a \neq 0,1$, and $b$ not rational and $c$ is transcendental.
- Substitute the exponential number $e$ for c: $$ a^b = e$$
- Raise both sides to the power of $(1/b)$ to obtain: $$a = e^{(1/b)}.$$
- Take the natural log of both sides: $$\ln(a) = \ln(e^{(1/b)}) = (1/b).$$
Assuming that $(1/b)$ is not rational, since b is not rational, this suggests that the natural logarithm of any algebraic number (irrational, rational or complex) $ \neq 0,1, $ is a non-rational algebraic number.
Hopefully, the issue of countable versus uncountable that flawed my last attempt is not an issue here.
Are there any counter examples?
You don't need to use the Gelfond-Schneider theorem. Just the transcendence of $e$ is enough.
To prove $r = \log a$ is irrational, suppose it is rational. Then $e^r = a$, so $e^r$ is algebraic. Since $a \not= 1$, $r \not= 0$.
Writing $r = m/n$ with integers $m$ and $n$, $(e^r)^n = e^{rn} = e^m$ is algebraic and $m \not= 0$.
Hermite showed is 1873 that $e$ is transcendental, so $e^m$ is transcendental when $m$ is a nonzero integer. Thus we have a contradiction, so $\log a$ is irrational.