I've seen this result usually given as "if $a$ and $b$ are coprime, then $\log\_{b}(a)$ is irrational." Wolfram|Alpha even says it is "unknown" whether or not $\log\_{12}(6)$ is irrational, but I think it is easy to show this is irrational (unless I'm missing something obvious):

By way of contradiction, suppose $\dfrac{\ln(6)}{\ln(12)} = \dfrac{n}{m}$ with integers $n,m$. Without loss of generality, take $n,m$ to be natural numbers.

Then $m\ln(6) = n\ln(12)$. Hence, $6^{m} = 12^{n}$, so we have $2^{m}3^{m} = 2^{2n}3^{n}$, and so:

$2^{m-2n}3^{m} = 3^{n}$

Since $3^{n}$ is odd, so is $2^{m-2n}3^{m}$, therefore $m-2n = 0$, and $m = 2n$. Hence $\dfrac{\ln(6)}{\ln(12)} = \dfrac{1}{2}$. This is a contradiction, so $\dfrac{\ln(6)}{\ln(12)}$ is irrational.

This leads me to believe that $a,b$ being coprime is a sufficient condition to say that $\log\_{b}(a)$ is irrational, but not a necessary one. My question is, can we give a stronger result regarding whether or not $\log\_{b}(a)$ is irrational? Can we give a necessary and sufficient condition? Does anyone know of a specific example of $a,b$ where the irrationality of $\log\_{b}(a)$ really is unknown?