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Some simplification gives: $2\log^2\left(\dfrac{x+4}{x}\right)=2\log^2\left(\dfrac{3-x}{x-1}\right)~(**)$ $\implies \log \left(\dfrac{x+4}{x}\right)=\pm\log\left(\dfrac{3-x}{x-1}\right)= \log\left(\dfrac{3-x}{x-1}\right)^{\pm 1}$ $\implies \dfrac{x+4}{x}= \dfrac{3-x}{x-1} \text{ and }\dfrac{x+4}{x}=\dfrac{x-1}{3-x}$

$\implies x^2-2=0\text{ and }x^2-6=0$

$\implies \large x=\sqrt 2,\sqrt 6\in(1,3)\small{\color{#3D99F6}{\text{ (See below) }}}$ i.e $\large \boxed 2$ values. $\therefore$ the correct option is $x^2-7x+10=0$ since $2$ is a root of this equation only.

Since domain of $\log$ function is $\mathbb{R^+}$ we get $\dfrac{x+4}{x}>0$ and $\dfrac{3-x}{x-1}>0$ . Solving we get $x\in(1,3)$ .