Rational exponential-logarithmic inequality

Problem

Solve the following inequality, writing a sign rule that also includes the existence conditions: $\frac{4-\left(\tfrac12\right)^{x-3}}{\big[\log_2 x-1\big]\big[\log_{1/2}(6-x)\big]}\ge 0.$

Solution

Existence conditions: $x>0$ (for $\log_2 x$), $6-x>0\Rightarrow x<6$, and the denominator factors nonzero: $\log_2 x-1\ne0\Rightarrow x\ne 2$, $\log_{1/2}(6-x)\ne0\Rightarrow 6-x\ne1\Rightarrow x\ne 5$. Domain: $(0,6)\setminus\{2,5\}$. Numerator $N=4-2^{3-x}$: $N\ge0\Leftrightarrow 2^{3-x}\le 4=2^{2}\Leftrightarrow 3-x\le 2\Leftrightarrow x\ge 1$. Denominator. $\log_2 x-1>0\Leftrightarrow x>2$; $\log_{1/2}(6-x)>0\Leftrightarrow 6-x<1\Leftrightarrow x>5$. So the product is positive on $(0,2)$ and $(5,6)$, negative on $(2,5)$. Sign table on $(0,6)$: the quotient is $\ge0$ where $N$ and $D$ have the same sign (or $N=0$). $\boxed{\,1\le x<2\ \ \lor\ \ 5<x<6\,}$