An "easy" way to get $\ln 7$ from $\ln 2$?

Question

I was trying to convince someone that there's no "easy formula" to get $\ln 7$ from $\ln 2$ (in the sense that it is easy to get $\ln 8$ by $3\ln 2$). I guess the formal way to state his question was whether $\ln 7$ is an element of $\mathbb{Q}(\ln 2)$. After some research, I now think that this is equivalent to $\ln 2$ and $\ln 7$ being algebraically independent over the rationals and that it is currently an open question whether that's the case or not. (In other words, if someone found such an "easy" formula, that would be a major surprise.) Is that correct or am I missing an easy proof?