I recently came across a number of the form $$ (\log\alpha)(\log\beta) $$ where $\alpha$ and $\beta$ were integers greater than 1. I wondered if the product was transcendental. Can this be proved? The natural next step would be $\alpha,\beta$ algebraic other than 0,1 (but I can’t even show the integer case so this is hopeless for me).
I was thinking in particular about A341577 but the problem is of course much more general, and I couldn't see a way to solve this with just the usual tools.
You can show unconditionally that the logarithms $\log p$ of the primes are linearly independent over $\mathbb{Q}$; this is equivalent to unique prime factorization. Given this, Schanuel's conjecture implies that the logarithms $\log p$ of the primes are in fact algebraically independent over $\mathbb{Q}$. This would imply in particular that for integers $n, m$, $\log n \log m$ is transcendental unless $n = 1$ or $m = 1$.
Unfortunately Schanuel's conjecture is, as far as I know, wide open.
If $\log(a)$ here means common log with base 10 then, no product of logs aren't transcendental, a counter example would be- Let's say $a=10$ and $b=100$ Then, $\log(a)=1$ and $\log(b)=2$ which are roots of $x^2-3x+2=0$
However if $\ln$ is taken log with base e then, $\ln(n)$ is transcendental for all $n>1$ where $n$ is non transcendental
Is ln(n) transcendental for all integer $n > 1$? You can get why $\ln(n)$ is transcendental here.
then, $\ln(a).\ln(b)$ has to be transcendental for all $a,b>1$ where $a,b$ are integers.As $\ln a$ and $\ln b$ are transcendental their product is as well.
(Product of $2$ transcendental not transcendental when the transcendental term gets cut like in $\frac{e^1}{e}$ one is the reciprocal of the other and $\ln x$ and $\frac{a}{\ln x}$ (a being any algebraic no.) don't intersect so not $\ln a$ not equal to $\frac{a}{b}$).
