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

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

Guest Feb 7, 2015

#7
+84167
+5

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)   ....

CPhill  Feb 7, 2015
Sort:

#1
+91928
+5

log on the base of sqrt ( 2 ) numbers ( x +4 ) is less than or equal to 2

$$\\\sqrt{x+4}\le 2\\\\ (\sqrt{x+4})^2\le 2^2\\\\ x+4\le 4\\\\ x+4-4\le 4-4\\\\ x\le 0\\\\$$

Melody  Feb 7, 2015
#2
+752
+3

what's the meaning of that question???

Sasini  Feb 7, 2015
#3
+11843
0

translator.......

i mean check in the translator!

rosala  Feb 7, 2015
#4
+91928
+5

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

$$\\x+4>0\\ x>-4\\\\ Hence\\\\ -4 or \\ alternatively the domain is  (-4,-2)$$
Melody  Feb 7, 2015
#5
+91928
+5

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\\\\

slight errors corrected - thanks Chris :))

Melody  Feb 7, 2015
#6
+91928
+5
Melody  Feb 7, 2015
#7
+84167
+5

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)   ....

CPhill  Feb 7, 2015
#8
+91928
0

Yes I like that Chris, thanks.

Melody  Feb 7, 2015

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