I was trying to figure out if $\ln(3)/\ln(2)$ is transcendental, when I found this post by b_jonas
But there's a proof just as simple showing that $\log 3/\log 2$ is
irrational. Suppose on contrary that $\log 3/\log 2 = p/q$ where p and q are
integers. Since $0< \log 3 / \log 2$ , we can choose p and q both as positive
integers. The equality then rearranges to $3^q = 2^p$ . But here, the left
hand side is odd and the right hand side is even, so we get a
contradiction.
that demonstrates the expression is not rational. With that clue it seems fairly easy to deduce that the expression is also transcendental. Here is how my argument goes:
Suppose there is a number $k$ such that $ 2^k = 3 \, $ then $\ln(2^k) = \ln(3) = \ln(2)*k \Rightarrow k =\ln(3)/ \ln(2) $. By the Gelfond–Schneider theorem $2^k $ should be transcendental if $k \neq 0,1$ is an algebraic irrational but $2^k = 3$, so k cannot be an algebraic irrational and it has already been demonstrated that $k$ is not an algebraic rational, so $\ln(3)/\ln(2)$ must be transcendental.
Is that correct?
If it is correct, it seems to follow that $ \ln(x) / \ln(y) $ is transcendental if $x$ and $y$ are integers with different parities, $(x,y \neq 0,1)$. (Later this extended to include the case where $x$ and $y$ are rational numbers with a few exceptions)
Best Answer
Looks correct to me, and I think it generalizes to $\ln(a)/\ln(b)$ being transcendental for any positive integers $a,b$ that admit no solution to $a^p=b^q$ for positive integers $p,q$.
Whether such a pair $p,q$ exists can be determined from the prime factorization of $a,b$. The pair exists if and only if the ratio between the powers of each prime is the same.
For example if $a=2^65^9$ and $b=2^45^6$ then since $6/4=9/6$ a pair exists, namely $a^2=b^3$.
If $a=3^65^8$ and $b=3^45^4$ then since $6/4\neq8/4$ no pair exists and $\ln(a)/\ln(b)$ is transcendental.