Linear independence of logs

Assume. $a\log2+b\log3=0$, so $\log(2^a3^b)=0$, i.e. $2^a3^b=1$.

Clear denominators. Multiply by a common multiple of denominators to get $2^m3^n=1$ with $m,n\in\mathbb{Z}$ not both zero.

Contradiction. If $m>0$, $2^m$ divides $3^n$, impossible by unique factorization. If $n>0$, similarly.

Conclusion. $\log2$ and $\log3$ are $\mathbb{Q}$-linearly independent; in particular $\log2/\log3$ is irrational.