$\begin{array}{l}\text { 10. (15 points) Given } a>0, b>0, \\ \log _{a} a=\log _{12} b=\log _{16}(a+b) . \\ \text { Find the value of } \frac{b}{a} .\end{array}$
10. Let $\log _{9} a=\log _{12} b=\log _{16}(a+b)=k$. Then $a=9^{k}, b=12^{k}, a+b=16^{k}$. Therefore, $(a+b) a=b^{2}$. Solving this, we get $\frac{b}{a}=\frac{1+\sqrt{5}}{2}$.