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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.

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2 Answers

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You got :

$$A^2+AB=B^2$$

Dividing by $AB$, we get :

$$\frac AB+1=\frac BA$$

Let $\dfrac AB =\color{gold}{x}$ and now solve quadratic equation in $\color{gold}x$.

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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, $$\frac{B}{A}=\frac{12^k}{9^k}=\left(\frac{4}{3}\right)^k=\left(\frac{4}{3}\right)^{\log_{\frac{3}{4}}\frac{\sqrt5-1}{2}}=\frac{2}{\sqrt5-1}=\frac{\sqrt5+1}{2}$$

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