Are logarithms of prime numbers quadratically independent over $\mathbb Q$?

Solution 1:

It seems highly probable that this is an open question, for the following indirect reason.

For non-negative integers $m_p, n_p$ ($p$ prime), almost all zero, if $a = \prod_pp^{m_p}$ and $b = \prod_pp^{n_p}$, then \begin{multline*} \log_2a = \log_3b \iff \frac{\sum_p m_p\log{p}}{\log2} = \frac{\sum_p n_p\log{p}}{\log3} \\ \iff -n_2(\log2)^2 + (m_2 - n_3)\log2\log3 + m_3(\log3)^2 \\ - \sum_{p\geqslant5}n_p\log2\log{p} + \sum_{p\geqslant5} m_p\log3\log{p} = 0, \end{multline*} and if the logarithms of the primes were known to be quadratically independent over $\mathbb{Q}$, this would imply $a = 2^n$, $b = 3^n$ for some non-negative integer $n$; but as this would settle the notorious open problem If $2^x $and $3^x$ are integers, must $x$ be as well?, someone would surely have noticed by now!