In sum of $n$, $S_n$ =$\frac{a(1-r^n)}{1-r}$, so if $r$ is positive and I wanted to find $n$. How do turn $r$ in to non-negative?

Question

I am given this question find the number of term in this geometric series $3 +1 + \frac{1}{3} , ... , \frac{121}{27}$. So $a=3$ and $r=\frac{1}{3}$ I subbed all the numbers in and I got $\frac{242}{243}-1=-\left(\frac{1}{3}\right)^n, (n = 5)$. I wanted to use logs to find n but I can't because $\frac{1}{3}$ is negative. So how do you do this question or am I using the right formula of $S_n =\frac{a(1-r^n)}{1-r}$ as $\frac{1}{3} <1$?