## College Algebra (10th Edition)

$3$
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: $\require{cancel} =\dfrac{3\cancel{\log{2}}}{\cancel{\log{2}}} \\=3$