8. The common ratio of the geometric sequence $a+\log _{2} 3, a+\log _{4} 3, a+\log _{8} 3$ is
8. $\frac{1}{3}$ Let the common ratio be $q$, then $q=\frac{a+\log _{4} 3}{a+\log _{2} 3}=\frac{a+\log _{8} 3}{a+\log _{4} 3}$, by the property of proportion we have: $q=\frac{a+\log _{4} 3-a-\log _{8} 3}{a+\log _{2} 3-a-\log _{4} 3}=$
$$
\frac{\log _{4} 3-\log _{8} 3}{\log _{2} 3-\log _{4} 3}=\frac{\log _{2} \sqrt{3}-\log _{2} \sqrt[3]{3}}{\log _{2} 3-\log _{2} \sqrt{3}}=\frac{\log _{2} \frac{\sqrt{3}}{\sqrt[3]{3}}}{\log _{2} \frac{3}{\sqrt{3}}}=\frac{\frac{1}{6} \log _{2} 3}{\frac{1}{2} \log _{2} 3}=\frac{1}{3}
$$