3. Solve the inequality $8^{\log _{2}^{2} x}-2 \cdot x^{\log _{2} x} \geqslant \log _{6} 108+\left(4-\log _{6}^{2} 3\right) \log _{108} 6$.
Answer. $x \in\left(0 ; \frac{1}{2}\right] \cup[2 ;+\infty)$.

Solution. Transform the right side of the inequality:

$\log _{6} 108+\left(4-\log _{6}^{2} 3\right) \log _{108} 6=2+\log _{6} 3+\frac{\left(2-\log _{6} 3\right)\left(2+\log _{6} 3\right)}{\log _{6} 108}=2+\log _{6} 3+\frac{\left(2-\log _{6} 3\right)\left(2+\log _{6} 3\right)}{2+\log _{6} 3}=4$.

Let $2^{\log _{2}^{2} x}=v$. Then $x^{\log _{2} x}=\left(2^{\log _{2} x}\right)^{\log _{2} x}=v$ and the inequality becomes $v^{3}-2 v-4 \geqslant 0$.

One of the roots of the polynomial on the left side is $v=2$. Factoring out the factor $v-2$, we get $(v-$ 2) $\left(v^{2}+2 v+2\right) \geqslant 0$, from which $v \geqslant 2$ (since $v>0$, the second factor is positive). We find $x$ :

$$
2^{\log _{2}^{2} x} \geqslant 2 \Leftrightarrow \log _{2}^{2} x \geqslant 1 \Leftrightarrow\left[\begin{array}{l}
\log _{2} x \geqslant 1 \\
\log _{2} x \leqslant-1
\end{array} \Leftrightarrow x \in\left(0 ; \frac{1}{2}\right] \cup[2 ;+\infty)\right.
$$