9. The common ratio of the geometric sequence $a+\log _{2} 3, a+\log _{4} 3, a+\log _{8} 3$ is
$$
\frac{1}{3}
$$
9. [Analysis and Solution] According to the problem, let's assume the common ratio is $q$, so we have $q=\frac{a+\log _{4} 3}{a+\log _{2} 3}=\frac{a+\log _{8} 3}{a+\log _{4} 3}$. According to the properties of ratios, we have
$$
q=\frac{a+\log _{4} 3-\left(a+\log _{8} 3\right)}{a+\log _{2} 3-\left(a+\log _{4} 3\right)}=\frac{\log _{4} 3-\log _{8} 3}{\log _{2} 3-\log _{4} 3}=\frac{\frac{1}{2} \log _{2} 3-\frac{1}{3} \log _{2} 3}{\log _{2} 3-\frac{1}{2} \log _{2} 3}=\frac{1}{3} .
$$