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

$2$
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$ $\therefore \log_3 8 = \dfrac{\log_{10} 8}{\log_{10} 3} \hspace{20pt} \text{and} \hspace{20pt} \log_8 9 = \dfrac{\log_{10} 9}{\log_{10} 8}$ Thus, the given exoression is equivalent to: $\require{cancel}\log_3 8\cdot \log_8 9 = \dfrac{\cancel{\log_{10} 8}}{\log_{10} 3} \cdot \dfrac{\log_{10} 9}{\cancel{\log_{10} 8}}= \dfrac{\log_{10} 9 }{\log_{10} 3}$ Note that: $\log_{10} 9 = \log_{10} 3^2$ $\because \log_a M^r = r \log_a M$ $\therefore \log_{10} 3^2 = 2 \log_{10} 3$ Therefore, $\require{cancel}\log_3 8\cdot \log_8 9 = \dfrac{2\cancel{ \log_{10} 3}}{\cancel{\log_{10} 3}} = 2$