2. Find all values of $y$, for each of which no value of $x$ satisfying the inequality $\log _{2}(|x|+|y|) \leqslant 2, \quad$ satisfies the inequality $\log _{\frac{1}{2}}(|x|+|y+4|) \geqslant-2$
Solution. The solutions of the first inequality on the Oxy plane form a square with vertices at points $(0 ; 4),(4 ; 0),(0 ;-4),(-4 ; 0)$, excluding the point with coordinates $(0 ; 0)$. The solution to the first inequality exists only if $y \in[-4 ; 4]$.

The solutions of the second inequality on the Oxy plane form a square with vertices at points $(0 ; 0),(4 ;-4),(0 ;-8)$,

![](https://cdn.mathpix.com/cropped/2024_05_06_9798297e598889c013ebg-13.jpg?height=217&width=380&top_left_y=2122&top_left_x=1158)

$(0,-4)$ vertices at points $(0 ; 0),(4 ;-4),(0 ;-8)$,

$(-4,-4)$

0 $\mid(0,-8)$ $(-4 ;-4)$, excluding the point with coordinates $(0 ;-4)$. The solution to the second inequality exists only if $y \in[-8 ; 0]$.

Therefore, for $y \in(-\infty ;-4] \cup[0 ;+\infty)$, no $x$ that is a solution to the first inequality is a solution to the second inequality.

Answer: $(-\infty ;-4] \cup[0 ;+\infty)$.

Answers in different versions: $(-\infty ;-4] \cup[0 ;+\infty) ;(-\infty ; 0] \cup[8 ;+\infty)$; $(-\infty ; 0] \cup[9 ;+\infty) ;(-\infty ;-8] \cup[0 ;+\infty)$.