Answer
$3A$
Work Step by Step
We are given that $log_{b}2=A$ and $log_{b}3=C$.
According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number).
Therefore, $log_{b}8=log_{b}(2^{3})=3log_{b}2=3(A)=3A$.