1. Given $a>0, b>0$, and $\log _{9} a=\log _{12} b=\log _{16}(a+b)$. Then the value of $\frac{b}{a}$ is $\qquad$ .
$\begin{array}{l}\text { - } 1 . \frac{1+\sqrt{5}}{2} . \\ \text { Given } \log _{9} a=\log _{12} b=\log _{16}(a+b)=k \\ \Rightarrow a=9^{k}, b=12^{k}, a+b=16^{k} \\ \Rightarrow \frac{b}{a}=\left(\frac{4}{3}\right)^{k}, 9^{k}+12^{k}=16^{k} \\ \Rightarrow\left(\frac{9}{12}\right)^{k}+1=\left(\frac{16}{12}\right)^{k} \\ \Rightarrow \frac{a}{b}+1=\frac{b}{a} \\ \Rightarrow\left(\frac{b}{a}\right)^{2}-\frac{b}{a}-1=0 \\ \Rightarrow \frac{b}{a}=\frac{1+\sqrt{5}}{2} .\end{array}$