Question

Find the set of real values of x for which ${\log }_{0.5}{\log }_2\frac{2x+1}{x+1}>0$

• A. x ∈ (-1,)
• B. x ∈ (1,)
• C. x ∈ (2,)
• D. x ∈ (0,)

Answer 1

The correct option is D

x ∈ (0,)

$\frac{2x+1}{x+1}$ > 0 $\Rightarrow (2x+1)(x+1)$ >0
$\Rightarrow$ x <$\frac{-1}{2}orx$ > -1 .........(1)
$lo{g}_2\left(\frac{2x+1}{x+1}\right)$ > 0
$\Rightarrow \frac{2x+1}{x+1}$ >1$\Rightarrow \frac{2x+1}{x+1}-1$ > 0
$\Rightarrow \frac{x}{x+1}$ >0
$\Rightarrow$ x < -1 or x > 0 ..........(2)
Now $lo{g}_{0.5}\left(lo{g}_2\right(\frac{2x+1}{x+1}\left)\right)$ >0
$\Rightarrow lo{g}_2\left(\frac{2x+1}{x+1}\right)$ <0
$\Rightarrow \frac{2x+1}{x+1}$< 2
$\Rightarrow \frac{2x+1}{x+1}-2$ <0
$\Rightarrow \frac{-1}{x+1}$ <0
$\Rightarrow x+1$> 0
$\Rightarrow$ x > -1 .......(3)
From (1),(2),(3), $xϵ(0,\infty )$