## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$3$
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$ $\therefore \log_2 6 = \dfrac{\log_{10} 6}{\log_{10} 2} \hspace{20pt} \text{and} \hspace{20pt} \log_6 8 = \dfrac{\log_{10} 8}{\log_{10} 6}$ Thus, the given expression is equivalent to: $\require{cancel} \log_2 6 \cdot \log_6 8 = \dfrac{\cancel{\log_{10} 6}}{\log_{10} 2} \times \dfrac{\log_{10} 8}{\cancel{\log_{10} 6}} = \dfrac{\log_{10} 8 }{\log_{10} 2}$ Note that: $\log_{10} 8 = \log_{10} 2^3$ $\because \log_a M^r = r \log_a M$ $\therefore \log_{10} 2^3 = 3 \log_{10} 2$ Therefore, $\require{cancel}\dfrac{\log_{10}{8}}{\log_{10}{2}} = \dfrac{3 \cancel{\log_{10} 2}}{\cancel{\log_{10} 2}} = 3$