Extreme logarithms problem

Let $\log_9 x = \log_{12} y = \log_{16} (x + y) = \alpha$ . Then $x = 9^{\alpha}, y = 12^{\alpha}$ and $x + y = 16^{\alpha}$ .

$\begin{array}{rll} x = \ 9^{\alpha} &\implies &\qquad \sqrt{x} = \ 3^{\alpha} \\ \ x + y = \ 16^{\alpha} &\implies &\ \ \sqrt{x + y} = \ 4^{\alpha} \end{array}$ $\begin{array}{rll} \dfrac{y}{x} = \dfrac{12^{\alpha}}{9^{\alpha}} &= \ \dfrac{4^{\alpha}}{3^{\alpha}} = \dfrac{\sqrt{x + y}}{\sqrt{x}} \\ \\ \dfrac{y^2}{x^2} &= \ \dfrac{x + y}{x} \\ \\ xy^2 - x^2y - x^3 &= \ 0 &\small \text{[We can divide by } x^3 \text{ since }x \ne 0] \\ \\ \dfrac{y^2}{x^2} - \dfrac{y}{x} - 1 &= \ 0 \\ \\ \dfrac{y}{x} &= \ \dfrac{1 \pm \sqrt{(\text{-}1)^2 - 4(1)(\text{-}1)}}{2(1)} &\small \text{[We can discard the negative solution as } \frac{y}{x} > 0] \\ \\ &= \ \dfrac{1 + \sqrt{5}}{2} \\ \\ &= \ \phi \approx \boxed{1.61803398875} \end{array}$

Let all these equal to $k$ . Then, $9^k=x,\space12^k=y,\space16^k=(x+y)$ $\space$ As, $16×9=12^2$ , $\implies\space x(x+y)=y^2$ Dividing throughout $x^2$ , $\frac {{y}^2}{ {x}^2}- \frac y x-1=0$ $\implies \frac y x= \frac{1+\sqrt5}{2}=\boxed{1.618}$

$\begin{aligned} \log_9 x & = \log_{12} y = \log_{16} (x+y) & \small \color{#3D99F6} \text{Convert to same logarithm base.} \\ \frac {\log x}{\log 9} & = \frac {\log y}{\log 12} = \frac {\log (x+y)}{\log 16} \end{aligned}$

$\implies \begin{cases} \dfrac {\log y}{\log x} = \dfrac {\log 12}{\log 9} = \dfrac {\log 3 + \log 4}{2 \log 3} = \dfrac 12 + \color{#D61F06}\dfrac {\log 4}{\log 9} & ... (1) \\ \dfrac {\log (x+y)}{\log x} = \dfrac {\log 16}{\log 9} = 2\cdot \color{#D61F06}\dfrac {\log 4}{\log 9} & ...(2) \end{cases}$

$\begin{aligned} (1): \quad \frac {\log y}{\log x} & = \frac 12 + \color{#D61F06} \frac {\log (x+y)}{2 \log x} & \small \color{#3D99F6} \text{Multiply both sides by }2 \log x. \\ 2 \log y & = \log x + \log(x+y) \\ \log y^2 & = \log (x(x+y)) \\ y^2 & = x(x+y) & \small \color{#3D99F6} \text{Divide both sides by }xy. \\ \frac yx & = \frac xy + 1 & \small \color{#3D99F6} \text{Multiply both sides by }\frac yx \text{ and rearrange.} \end{aligned}$

$\begin{aligned} \left(\frac yx\right)^2 - \frac yx - 1 & = 0 & \small \color{#3D99F6} \text{Solve the quadratic for } \frac yx. \\ \implies \frac yx & = \frac {1+\sqrt 5}2 & \small \color{#3D99F6} \text{Note that }\frac yx > 0 \\ & \approx \boxed{1.618} \end{aligned}$