Answer
$2$
Work Step by Step
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$
$\therefore \log_3 8 = \dfrac{\log_{10} 8}{\log_{10} 3} \hspace{20pt} \text{and} \hspace{20pt} \log_8 9 = \dfrac{\log_{10} 9}{\log_{10} 8}$
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,
$\require{cancel}\log_3 8\cdot \log_8 9 = \dfrac{2\cancel{ \log_{10} 3}}{\cancel{\log_{10} 3}} = 2$
