5. Solve the inequality

$$
\frac{\left(5 \cdot 2^{-\log _{x} 3}-2.5\right) \sqrt{2-\sqrt{\log _{x} 3+1}}}{1+\sqrt{\log _{x} 3+8}}>\frac{\left(2^{-\log _{x} 3}-0.5\right) \sqrt{2-\sqrt{\log _{x} 3+1}}}{\sqrt{\log _{x} 3+8}-3}
$$
Solution: Let's make the substitution $y=\log _{x} 3$.

$$
\frac{\left(5 \cdot 2^{-y}-2.5\right) \sqrt{2-\sqrt{y+1}}}{1+\sqrt{y+8}}>\frac{\left(2^{-y}-0.5\right) \sqrt{2-\sqrt{y+1}}}{\sqrt{y+8}-3}
$$

$\left(2^{-y}-0.5\right)\left(\frac{5}{1+\sqrt{y+8}}-\frac{1}{\sqrt{y+8}-3}\right) \sqrt{2-\sqrt{y+1}}>0$ Find the domain: $\quad y \geq-8, \quad y \neq 1, \quad 2-\sqrt{y+1} \geq 0$ $\Rightarrow-1 \leq y \leq 3$. Since the inequality is strict, then for $-1 \leq y<1$, or $(y-1)\left(\frac{4 \sqrt{y+8}-16}{(1+\sqrt{y+8})(\sqrt{y+8}-3)}\right) > 0 , } \\
{ \log _{3} x \leq -1 , } \\
{ \log _{3} x > 1 / 3 , } \\
{ x \neq 3 , }
\end{array} }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
x \in(0 ; 1 / 3] \cup(1 ;+\infty), \\
x \in(0 ; 1) \cup(\sqrt[3]{3} ;+\infty), \\
x \neq 3,
\end{array} \Leftrightarrow\right.\right.\right.\right.
$$

$x \in(0 ; 1 / 3] \cup(\sqrt[3]{3} ; 3) \cup(3 ;+\infty)$.

Answer: $x \in(0 ; 1 / 3] \cup(\sqrt[3]{3} ; 3) \cup(3 ;+\infty)$.