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Let $n_1, m_1, n_2, m_2 \in \mathbb{N}_{\geq 2}$ such that $\gcd(n_1, m_1) = \gcd(n_2, m_2) = 1$ and $$ \log_{n_1}(m_1) = \log_{n_2}(m_2). $$ Does it follow that $n_1 = n_2$ and $m_1 = m_2$?

Equivalent formulation

This is the original motivation. Let $n_1, m_1, n_2, m_2 \in \mathbb{N}_{\geq 2}$ such that $\gcd(n_1, m_1) = \gcd(n_2, m_2) = 1$. Can we always find $a, b \in \mathbb{Z}$ such that $$ n_1^a m_1^b > 1 \\ n_2^a m_2^b < 1 ? $$

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The claim does not follow. Consider: $$ \log_4 9=\log_2 3. $$

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  • $\begingroup$ You are right, I asked the wrong question. I really should have asked about squarefree $n_1, n_2, m_1, m_2$, but I got greedy. $\endgroup$ – Vincent Mar 21 at 16:08
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    $\begingroup$ @Vincent With squarefree $n_1,n_2,m_1,m_2$ it should work. Probably even much weaker restriction that $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$ are not powers of the same rational number would suffice. $\endgroup$ – user Mar 21 at 17:02

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