Answer
$3$
Work Step by Step
$\because \log_a M = \dfrac{\log_b M}{\log_b a}$
$\therefore \log_2 6 = \dfrac{\log_{10} 6}{\log_{10} 2} \hspace{20pt} \text{and} \hspace{20pt} \log_6 8 = \dfrac{\log_{10} 8}{\log_{10} 6}$
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,
$\require{cancel}\dfrac{\log_{10}{8}}{\log_{10}{2}} = \dfrac{3 \cancel{\log_{10} 2}}{\cancel{\log_{10} 2}} = 3$
