[tex]\\ $Punem conditiile de existenta: \\ \\ x $ \ \textgreater \ 0 \quad $si$ \quad y\ \textgreater \ 0 \Rightarrow \dfrac{x}{y} \ \textgreater \ 0.[/tex]
[tex]2 lnx - lny = ln\Big(x+\dfrac{3}{4}y\Big) \Rightarrow lnx^2 - lny = ln\Big(x+\dfrac{3}{4}y\Big) \Rightarrow \\ \\ \Rightarrow ln\Big(\dfrac{x^2}{y}\Big) = ln\Big(x+\dfrac{3}{4}y\Big) \Rightarrow \dfrac{x^2}{y} = x+\dfrac{3}{4}y \Big|:x \Rightarrow \dfrac{x}{y} = 1+\dfrac{3}{4x}y \Rightarrow \\ \\ \Rightarrow \dfrac{x}{y} = 1 +\dfrac{3y}{4x} \Rightarrow \dfrac{x}{y} = 1 + \dfrac{3}{4}\cdot \dfrac{y}{x} \Rightarrow \dfrac{x}{y} = 1 +\dfrac{3}{4} \cdot \dfrac{1}{\dfrac{x}{y}} [/tex]
[tex]\\ \\ $Notam: \dfrac{x}{y} =t \\ \\ \Rightarrow t=1+\dfrac{3}{4}\cdot \dfrac{1}{t} \Big|\cdot t \Rightarrow t^2=t+\dfrac{3}{4} \Big|\cdot 4 \Rightarrow 4t^2 = 4t+3 \Rightarrow \\ \\ \Rightarrow 4t^2-4t-3 = 0 \\ \\ \Delta = (-4)^2 - 4\cdot 4\cdot (-3) = 16+16\cdot 3 = 16+48 = 64 \\ \\ \Rightarrow t_{1,2} = \dfrac{4\pm \sqrt{64}}{2\cdot 4} = \dfrac{4\pm8}{2\cdot 4} = \dfrac{1\pm2}{2} \Rightarrow t_1 = -\dfrac{1}{2},\quad t_2 = \dfrac{3}{2}; \\ [/tex]
[tex]\Rightarrow t\in \Big\{-\dfrac{1}{2},\dfrac{3}{2}\Big\} \Rightarrow \dfrac{x}{y}\in \Big\{-\dfrac{1}{2},\dfrac{3}{2}\Big\} \\ \\ $Dar \dfrac{x}{y} \ \textgreater \ 0 \Rightarrow \boxed{\dfrac{x}{y} = \dfrac{3}{2}} \rightarrow $ c) corect$[/tex]
