## College Algebra (10th Edition)

$3$
RECALL: The change-of-base formula for logarithms: $\log_a{b} = \dfrac{\log{b}}{\log{a}}$ Use the change-of-base formula above: $=\dfrac{\log{4}}{\log{2}} \cdot \dfrac{\log{6}}{\log{4}} \cdot \dfrac{\log{8}}{\log{6}}$ Cancel the common factors: $\require{cancel} =\dfrac{\cancel{\log{4}}}{\log{2}} \cdot \dfrac{\cancel{\log{6}}}{\cancel{\log{4}}} \cdot \dfrac{\log{8}}{\cancel{\log{6}}} \\=\dfrac{\log{8}}{\log{2}} \\=\dfrac{\log{2^3}}{\log{2}}$ Use the rule $\log{(M^r)}=r \cdot \log{M}$ to obtain: $=\dfrac{3\cdot\log{2}}{\log{2}}$ Cancel common factors to obtain: $\require{cancel} =\dfrac{3\cancel{\log{2}}}{\cancel{\log{2}}} \\=3$