8. Let there be a geometric sequence $\left(a+\log _{2} 3\right),\left(a+\log _{4} 3\right),\left(a+\log _{8} 3\right)$, with a common ratio of $\qquad$ .
8. $\frac{1}{3}$.

According to the problem, we have $\left(a+\log _{2} 3\right)\left(a+\log _{8} 3\right)=\left(a+\log _{4} 3\right)^{2}$. Let $\log _{2} 3=m$, then $\log _{8} 3=\frac{1}{3} m, \log _{4} 3=\frac{1}{2} m, a=-\frac{m}{4}$. The three terms are $\frac{3}{4} m, \frac{1}{4} m, \frac{1}{12} m$, so the common ratio is $\frac{1}{3}$.