3. [5 points] Solve the inequality $8^{\sqrt{\log _{2} x}}-7 \cdot 2^{1+\sqrt{4 \log _{2} x}}+60 \cdot x \sqrt{\log _{x} 2} \leqslant 72$.
Answer: $x \in(1 ; 2] \cup\left\{6^{\log _{2} 6}\right\}$.

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 $2^{3 \sqrt{\log _{2} x}}-14 \cdot 2^{2 \sqrt{\log _{2} x}}+60 \cdot 2^{\sqrt{\log _{2} x}} \leqslant 72$. (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 $t=2 \sqrt{\log _{2} x}$. Then we get $t^{3}-14 t^{2}+60 t-72 \leqslant 0$, or, factoring the left side, $(t-2)(t-6)^{2} \leqslant 0$, from which it follows that $t \in(-\infty ; 2] \cup\{6\}$.

Returning to the variable $x$, we find that

$$
\left[\begin{array}{l}
2^{\sqrt{\log _{2} x}} \leqslant 2, \\
2^{\sqrt{\log _{2} x}} = 6
\end{array} \Leftrightarrow \left[\begin{array}{l}
\sqrt{\log _{2} x} \leqslant 1 \\
\sqrt{\log _{2} x} = \log _{2} 6
\end{array} \Leftrightarrow \left[\begin{array}{l}
0 \leqslant \log _{2} x \leqslant 1 \\
\log _{2} x = \log _{2}^{2} 6
\end{array} \Leftrightarrow \left[\begin{array}{l}
1 \leqslant x \leqslant 2 \\
x=6^{\log _{2} 6}
\end{array}\right.\right.\right.\right.
$$

It was obtained above that $x \neq 1$, so the final result is: $x \in(1 ; 2] \cup\left\{6^{\log _{2} 6}\right\}$.