Example 23. Solve the system

$$
\left\{\begin{array}{l}
4 \log _{2}^{2} x+1=2 \log _{2} y \\
\log _{2} x^{2} \geqslant \log _{2} y
\end{array}\right.
$$
Solution. The domain of admissible values of the system is defined by the system of inequalities $x>0, y>0$. The second inequality of the system on the domain of admissible values is equivalent to the inequality $2 \log _{2} x \geqslant \log _{2} y$, replacing $2 \log _{2} y$ in which with $4 \log _{2}^{2} x+1$, we obtain the inequality

$$
4 \log _{2}^{2} x-4 \log _{2} x+1 \leqslant 0
$$

i.e.

$$
\left(2 \log _{2} x-1\right)^{2} \leqslant 0
$$

This inequality holds if and only if $2 \log _{2} x-1=0$, i.e., when $x=\sqrt{2}$.

For $x=\sqrt{2}$, from the equation of the system we find $y=2$.

Verification shows that the pair of numbers $(\sqrt{2} ; 2)$ satisfies the given system, and, therefore, the original system has a unique solution $(\sqrt{2} ; 2)$.