1.4.12 $\star \star$ Solve the following inequality
$$
\cdot\left|\log _{2} x-3\right|+\left|2^{x}-8\right| \geqslant 9 \text {. }
$$
Given the condition $x>0$, we discuss in three cases.

(1) When $x \in(0,3]$, the inequality (1) becomes $3-\log _{2} x+8-2^{x} \geqslant 9$, which simplifies to $2^{x}+\log _{2} x \leqslant 2$. Since $f(x)=2^{x}+\log _{2} x$ is increasing on $(0,3]$, and given $f(1)=2$, we can conclude that $0<x \leqslant 1$. Therefore, the solution set for this case is $(0,1]$.

(2) When $x \in(3,8]$, the inequality (1) becomes $2^{x}-\log _{2} x \geqslant 9$. Let $g(x)=2^{x}-\log _{2} x$. Since $g'(x)=2^{x} \ln 2 - \frac{1}{x \ln 2}>0$, $g(x)$ is increasing for $x \geqslant 3$. Also, $g(4)=2^{4}-\log _{2} 4=14$, so the solution set for this case is $[4,8]$.

(3) When $x>8$, the inequality (1) becomes $\log _{2} x-3+2^{x}-8 \geqslant 9$, which simplifies to $2^{x}+\log _{2} x \geqslant 20$. For $x>8$, $2^{x}+\log _{2} x>2^{8}+\log _{2} 8=259>20$, so $x>8$ satisfies the inequality (1). Therefore, the solution set for the inequality is $(0,1] \cup[4,+\infty)$.