3. Let $f(x)=1+\log _{x} 5, g(x)=\log _{x}^{2} 9+\log _{x}^{3} 8$, compare the size of $f(x)$ and $g(x)$.
(1985 Shanghai Mathematics Competition Problem)
3. Since $f(x)=1+\log _{x} 5, g(x)=\log _{x}^{2} 9+\log _{x}^{3} 8=\log _{x} 3+\log _{x} 2=\log _{x} 6$, therefore $f(x)-g(x)=1-\log _{x} \frac{6}{5}$, thus, when $x=\frac{6}{5}$, $f(x)=g(x)$; when $1<x<\frac{6}{5}$, $f(x)>g(x)$; when $x>\frac{6}{5}$, $f(x)<g(x)$.