First, notice that for any positive rational number $a$, $(a, 1)$ is a solution. Now, suppose $b \neq 1$. We prove first the following: Lemma: If $0<p< q$ are positive integers such that $\gcd(p,q)=1$ and $0<x<1$ is a rational number such that $x^{q-p}=\frac{p}{q}$, then $q=p+1$. Proof: Write $x=\frac{u}{v}$ where $\gcd(u, v)=1$ and $0<u<v$. Let $d=q-p$, then $v^d-u^d=d$, but $v^d-u^d \geq (u+1)^d - u^d=\sum_{i=0}^{d-1} {d \choose i} u^i \geq d$ with equality if and only if $d=1$. Now, write $a= \prod_{i}p_i ^ {\alpha_i}$ and $b=\prod_{i}p_i ^ {\beta_i}$, where $\alpha_i, \beta_i \in \mathbb{Z}$. Write $b=\frac{p}{q}$ where $p, q$ are positive integers s.t. $\gcd(p,q)=1$. The equation implies $q\alpha_i=p(\alpha_i+\beta_i)$, therefore $\alpha_i=p \gamma_i$ for some $\gamma_i \in \mathbb{Z}$, hence $\beta_i = (q-p) \gamma_i$. Let $x=\prod_{i}p_i ^ {\gamma_i}$, then $a=x^p$ and $b=x^{q-p}=\frac{p}{q}$. Case 1: $q > p$ Using the lemma, we get $b=\frac{p}{p+1}$ and $a=\left ( \frac{p}{p+1}\right ) ^ p$, which is a solution whenever $p>0$ is an integer. Case 1: $q < p$ Interchange $p$ and $q$ and apply the lemma to get $b=\frac{q+1}{q}$ and $a=\left ( \frac{q}{q+1}\right ) ^ {q+1}$, which is a solution whenever $q>0$ is an integer. Conclusion: The solution are: $(a, 1)$ for any rational number $a>0$. $\left (\left ( \frac{p}{p+1}\right ) ^ p, \frac{p}{p+1} \right)$ for any positive integer $p$. $\left ( \left ( \frac{p}{p+1}\right ) ^ {p+1}, \frac{p+1}{p} \right)$ for any positive integer $p$.