3. [5 points] Solve the inequality $27^{\sqrt{\log _{3} x}}-13 \cdot 3 \sqrt{\sqrt{4 \log _{3} x}}+55 \cdot x^{\sqrt{\log _{x} 3}} \leqslant 75$.
Answer: $x \in(1 ; 3] \cup\left\{5^{\log _{3} 5}\right\}$.

Solution. Note that $x \sqrt{\sqrt{\log _{x} 3}}=\left(3^{\log _{3} x}\right)^{\sqrt{\log _{x} 3}}=3^{\sqrt{\log _{3}^{2} x \cdot \log _{x} 3}}=3 \sqrt{\log _{3} x}$. Therefore, the inequality can be rewritten as $3^{3 \sqrt{\log _{3} x}}-13 \cdot 3^{2 \sqrt{\log _{3} x}}+55 \cdot 3^{\sqrt{\log _{3} x}} \leqslant 75$. (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 $t=3 \sqrt{\log _{3} x}$. Then we get $t^{3}-13 t^{2}+55 t-75 \leqslant 0$, or, factoring the left side, $(t-3)(t-5)^{2} \leqslant 0$, from which it follows that $t \in(-\infty ; 3] \cup\{5\}$.

Returning to the variable $x$, we find that

$$
\left[\begin{array}{l}
3^{\sqrt{\log _{3} x}} \leqslant 3, \\
3 \sqrt{\log _{3} x} = 5
\end{array} \Leftrightarrow \left[\begin{array}{l}
\sqrt{\log _{3} x} \leqslant 1 \\
\sqrt{\log _{3} x} = \log _{3} 5
\end{array} \Leftrightarrow \left[\begin{array}{l}
0 \leqslant \log _{3} x \leqslant 1 \\
\log _{3} x = \log _{3}^{2} 5
\end{array} \Leftrightarrow \left[\begin{array}{l}
1 \leqslant x \leqslant 3 \\
x=5^{\log _{3} 5}
\end{array}\right.\right.\right.\right.
$$

It was obtained above that $x \neq 1$, so the final result is: $x \in(1 ; 3] \cup\left\{5^{\log _{3} 5}\right\}$.