LHS $=\log_{2}[(x^{2}-3x+2)(x^{2}-7x+12)]= \log_2[(x-1)(x-2)(x-3)(x-4)].$

RHS $=\log_2 8 + \log_2\!\left[\dfrac{(x-1)(x-4)}{2}\right] = \log_2\!\left[\dfrac{8(x-1)(x-4)}{2}\right] = \log_2[4(x-1)(x-4)].$

Domain: $(x-1)(x-2)>0,\ (x-3)(x-4)>0,\ (x-1)(x-4)>0.$ Intersection: $x<1$ or $x>4.$

On this domain, $(x-1)(x-4)>0$, so the equation becomes

$

(x-1)(x-2)(x-3)(x-4) = 4(x-1)(x-4) ;\Longrightarrow; (x-2)(x-3) = 4.

$

$x^{2}-5x+6=4 \;\Longrightarrow\; x^{2}-5x+2=0 \;\Longrightarrow\; x = \dfrac{5\pm\sqrt{17}}{2}.$

$\dfrac{5+\sqrt{17}}{2}\approx 4.56 > 4$ \checkmark and $\dfrac{5-\sqrt{17}}{2}\approx 0.44 < 1$ \checkmark. Both are in the domain.

So there are 2 real values.

Why this is CAT-level: The temptation is to immediately set $(x-2)(x-3)=4$ without checking the log domain. Students who skip the domain analysis still arrive at $2$ -- but by accident. A natural error path is to assume one of the roots is rejected by the domain; only careful interval analysis confirms both lie in the valid region.

Answer: (C) $2$.