Log of A to the base 9=Log of B to the base 12=Log of (A+B) to the base of 16, then what is B/A? I tried, 
$9^k$=A, $12^k$=B, $16^k$=A+B 
Now, $16^k$*$9^k$=$(12^k$)^2
That means, (A+B)*A=$B^2$,
therefore, $A^2$-$B^2$=-AB
and, (A-B)(A+B)=-AB
and, (B-A)(B+A)=AB.
I don't know what to do further. 

Let $\log_9A=\log_{12}B=\log_{16}(a+B)=k$.
Hence, $A=9^k$, $B=12^k$ and $A+B=16^k$, which gives
$$\left(3^k\right)^2+3^k\cdot4^k-\left(4^k\right)^2=0$$ or
$$\left(\frac{3}{4}\right)^k=\frac{\sqrt5-1}{2}.$$
Thus,
$$\frac{B}{A}=\frac{12^k}{9^k}=\left(\frac{4}{3}\right)^k=\left(\frac{4}{3}\right)^{\log_{\frac{3}{4}}\frac{\sqrt5-1}{2}}=\frac{2}{\sqrt5-1}=\frac{\sqrt5+1}{2}$$
