Independent over rational number field

Question

Is there some theorems which can make sure that $$1, \frac{\log 2}{\log 3}, \frac{\log 3}{\log 2}$$ are $\mathbb{Q}$-independent?

Answer 1

$\def\Q{\mathbb{Q}}$ Edit: Suppose $$1, \frac{\log 2}{\log 3}, \frac{\log 3}{\log 2}$$ are $\Q$-dependent. That means that there exists $a,b,c\in\Q$, not all zero, such that \begin{equation} b\cdot1+a\frac{\log 2}{\log 3}+c\frac{\log 3}{\log 2}=0. \end{equation} Without loss of generality, we can suppose $a,b,c$ to be integers, and $a>0$. If $a=0$, then we easily get a contradiction. Now, set $$ x=\frac{\log 2}{\log 3}. $$ Observe that $$ x=\log_3 2. $$ Now we have $$ b+ax+cx^{-1}=0, $$ so $$ ax^2+bx+c=0. $$ But $x=\log_3 2$ is notoriously trascendental (see $\log_3 2$ is trascendental), contradiction. Therefore $$1, \frac{\log 2}{\log 3}, \frac{\log 3}{\log 2}$$ are $\Q$-independent.