Answer
$3$
Work Step by Step
We know that $\log_a {b}=\dfrac{\log_c {b}}{\log_c {a}}$.
Hence,
$\log_6 {8}=\dfrac{\log_2 {8}}{\log_2 {6}}$.
Thus,
$\log_2 {6}\cdot \log_6 {8}
\\=\log_2 {6}\cdot \dfrac{\log_2 {8}}{\log_2 {6}}
\\=\log_2 {8}
\\=\log_2 {2^3}$
Use the rule $\log_a{a^x}=x$ to obtain:
$\log_2{2^3} = 3$
