Answer
$3$
Work Step by Step
We know that $\log_a {b}=\frac{\log_c {b}}{\log_c {a}}$ (this is known as the Change-of-Base formula), hence:
$\log_2 {4}\cdot \log_4 {6}\cdot \log_6 {8}=$
Cancel the common factors to obtain
$=\frac{\log{4}}{\log{2}}\cdot\frac{\log{6}}{\log{4}}\cdot\frac{\log{8}}{\log{6}}=\frac{\log{8}}{\log{2}}$
By the Change-of-Base formula: $\frac{\log{8}}{\log{2}}=\log_2 {8}=\log_2 {2^3}.$
We know that $\log_a {x^n}=n\cdot \log_a {x}$, hence $\log_2 {2^3}=3\log_2{2}=3\cdot1=3.$
