Answer
$log(r)$, see explanations.
Work Step by Step
Step 1. Given $a_n$ as a geometric sequence with $r\gt0, a_1\gt0$, we have $a_n\gt0$ and $\frac{a_{n+1}}{a_n}=r$
Step 2. Examine the sequence $b_n=log(a_n)$, we have $b_{n+1}-b_n=log(a_{n+1})-log(a_n)=log\frac{a_{n+1}}{a_n}=log(r)$
Step 3. Thus, sequence $b_n=log(a_n)$ is of arithmetic with a common difference of $log(r)$
