Answer
$3$
Work Step by Step
RECALL:
(1)The change-of-base formula for logarithms:
$\log_a{b} = \dfrac{\log{b}}{\log{a}}$
(2) $\log_a{(B^n)}=n\cdot\log_a{B}$
Use rule (1) above to obtain:
$\log_2{6} \cdot \log_6{8}=\dfrac{\log{6}}{\log{2}} \cdot
\dfrac{\log{8}}{\log{6}}$
Cancel the common factors to obtain:
$\require{cancel}
=\dfrac{\cancel{\log{6}}}{\log{2}} \cdot
\dfrac{\log{8}}{\cancel{\log{6}}}
\\=\dfrac{\log{8}}{\log{2}}
\\=\dfrac{\log{2^3}}{\log{2}}$
Use rule (2) above to obtain:
$=\dfrac{3\log{2}}{\log{2}}$
Cancel the common factors to obtain:
$\require{cancel}
=\dfrac{3\cancel{\log{2}}}{\cancel{\log{2}}}
\\=3$
