3. [5 points] Solve the inequality $8^{\sqrt{\log _{2} x}}-2^{\sqrt{4 \log _{2} x}+3}+21 \cdot x^{\sqrt{\log _{x} 2}} \leqslant 18$.
Answer: $x \in(1 ; 2] \cup\left\{3^{\log _{2} 3}\right\}$.

Solution. Note that $x^{\sqrt{\log _{x} 2}}=\left(2^{\log _{2} x}\right)^{\sqrt{\log _{x} 2}}=2^{\sqrt{\log _{2}^{2} x \cdot \log _{x} 2}}=2^{\sqrt{\log _{2} x}}$. Therefore, the inequality can be rewritten as $2^{3 \sqrt{\log _{2} x}}-8 \cdot 2^{2 \sqrt{\log _{2} x}}+21 \cdot 2^{\sqrt{\log _{2} x}} \leqslant 18$. (Note that the domain of permissible values has changed: the restriction $x \neq 1$ has disappeared, which will need to be taken into account later.) Let $w=2^{\sqrt{\log _{2} x}}$. Then we get $w^{3}-8 w^{2}+21 w-18 \leqslant 0$, or, factoring the left side, $(w-2)(w-3)^{2} \leqslant 0$, from which it follows that $w \in$ $(-\infty ; 2] \cup\{3\}$.

Returning to the variable $x$, we find that

$$
\left[\begin{array}{l}
2^{\sqrt{\log _{2} x}} \leqslant 2, \\
2^{\sqrt{\log _{2} x}} = 3
\end{array} \Leftrightarrow \left[\begin{array}{l}
\sqrt{\log _{2} x} \leqslant 1, \\
\sqrt{\log _{2} x} = \log _{2} 3
\end{array} \Leftrightarrow \left[\begin{array}{l}
0 \leqslant \log _{2} x \leqslant 1 \\
\log _{2} x = \log _{2}^{2} 3
\end{array} \Leftrightarrow \left[\begin{array}{l}
1 \leqslant x \leqslant 2 \\
x=3^{\log _{2} 3}
\end{array}\right.\right.\right.\right.
$$

It was obtained above that $x \neq 1$, so the final result is: $x \in(1 ; 2] \cup\left\{3^{\log _{2} 3}\right\}$.