# Are there nontrivial natural number solutions for $\log_{n_1}(m_1) = \log_{n_2}(m_2)$ with $\gcd(n_1, m_1) = \gcd(n_2, m_2) = 1$?

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 $$n_1^a m_1^b > 1 \\ n_2^a m_2^b < 1 ?$$

The claim does not follow. Consider: $$\log_4 9=\log_2 3.$$
• 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. – Vincent Mar 21 at 16:08
• @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. – user Mar 21 at 17:02