1. If positive numbers $a, b$ satisfy $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$, then the value of $\frac{1}{a}+\frac{1}{b}$ is
Answer: 108.
Solution: Let $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$, then $a=2^{k-2}, b=3^{k-3}, a+b=6^{k}$, thus $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}=\frac{6^{k}}{2^{k-2} \times 3^{k-3}}=2^{2} \times 3^{3}=108$