Question 22: Positive real numbers $a$ and $b$ satisfy $\log _{9} a=\log _{12} b=\log _{16}(3 a+b)$, then $\frac{b}{a}=$ $\qquad$
Question 22, Answer: Let $\log _{9} a=\log _{12} b=\log _{16}(3 a+b)=t$, then $\left\{\begin{array}{c}a=9^{t}=3^{2 t} \\ b=12^{t} \\ 3 a+b=16^{t}=4^{2 t}\end{array}\right.$, so $a(3 a+b)=b^{2} \Rightarrow \frac{b}{a}=\frac{1+\sqrt{13}}{2}$.