Помогите пожалуйста..

Решите задачу:

$log_3log_\frac{9}{16}(x^2-4x+3)=0\\OD3:x^2-4x+3\ \textgreater \ 0\\\begin{cases}x_1+x_2=4\\x_1*x_2=3\end{cases}\\x_1=1\ ;x_2=3\\///(+)///(1)...(-)...(3)///(+)///\ \textgreater \ x\\x\in(-\infty;1)\cup(3;+\infty)\\log_\frac{9}{16}(x^2-4x+3)=1\\\frac{1}{2}log_\frac{3}{4}(x^2-4x+3)=1\\\sqrt{x^2-4x+3}=\frac{3}{4}\\OD3:x^2-4x+3\ \textgreater \ 0\\\begin{cases}x_1+x_2=4\\x_1*x_2=3\end{cases}\\x_1=1\ ;x_2=3\\///(+)///(1)...(-)...(3)///(+)///\ \textgreater \ x\\x\in(-\infty;1)\cup(3;+\infty)$
$x^2-4x+3=\frac{9}{16}\\x^2-4x+\frac{39}{16}=0\\16x^2-64x+39=0\\x_{1,2}=\frac{32^+_-\sqrt{400}}{16}=\frac{32^+_-20}{16}\\x_1=\frac{13}{4}\ \ ;x_2=\frac{3}{4}$