Question

Two positive real numbers $a$ and $b$ satisfy $2+{\log }_{2}a=3+{\log }_{3}b={\log }_6(a+b)$, then value of $\frac{1}{a}+\frac{1}{b}$ is

Answer 1

$\text{Let}2+{\log }_{2}a=3+{\log }_{3}b={\log }_6(a+b)=k$

$a=2^{k-2},b=3^{k-3},a+b=6^k$
$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{6^k}{2^{k-2}{.3}^{k-3}}=2^2.3^3$
$=4\times 27=108$