You know Sasini, I do not think I answered the intended question.
I might try it again. (It is written in Russian)
I use Bing or Google translators to make sense of these.
logonthe base ofsqrt (2)numbers (x+4)is less than or equal to2
$$\\log_{\sqrt{2}}\;(x+4)\le 2\\\\
\sqrt{2}^{(log_{\sqrt{2}}\;(x+4))}\le\sqrt{2}^ 2\\\\
(x+4)\le 2\\\\
x\le -2\\\\$$
But you can only find the log of a positive number so
$$\\x+4>0\\
x>-4\\\\
Hence\\\\
-4 or \\
$alternatively the domain is $ (-4,-2)$$
.