What is the exact value of $2^{\log _{6} 18} \cdot 3^{\log _{6} 3}$?
Apply the known identities of logarithms and exponentiation, as well as the definition of logarithms:

$$
2^{\log _{6} 18}=2^{\log _{6} 3 \cdot 6}=2^{\log _{6} 3+\log _{6} 6}=2^{\log _{6} 3+1}=2 \cdot 2^{\log _{6} 3}
$$

Substituting this into (1), we get

$$
2 \cdot 2^{\log _{6} 3} \cdot 3^{\log _{6} 3}=2 \cdot 6^{\log _{6} 3}=2 \cdot 3=6
$$