log по основанию sqrt(2) числа (x+4) меньше или равно 2

Here's another way

log_√2(x + 4) ≤ 2

This says that

(√2)^2 ≥ x + 4    and, remembering that x + 4 > 0   ....so we have

2 ≥ x + 4          and      x + 4 > 0

-2 ≥ x               and      x > -4

So, the solution is

(-4, -2]  for x   ...note that -2 is included because (√2)^2  ≤ (-2 + 4)   ....

what's the meaning of that question???

translator.......

i mean check in the translator!

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.

I might try looking at this question a more straight forward way.

$$\\NOTE: \;\;x+4>0\;\;so\;\;x>-4\\\\
log_{\sqrt2}(x+4)\le2\\\\
\frac{log(x+4)}{log{\sqrt2}}\le2\\\\
log(x+4)\le 2log\sqrt2\\\\
log(x+4)\le 2log(2^{1/2})\\\\
log(x+4)\le log(2)\\\\
x+4\le2\\\\
x \le -2\\\\
therefore\qquad -4

slight errors corrected - thanks Chris :))

Here is the Wolfram|Alpha solution and  graph

http://www.wolframalpha.com/input/?i=log%28x%2B4%29%2Flog%28sqrt2%29%3C2

Yes I like that Chris, thanks.