Question

If ${\log }_{0.5}\left(\frac{3-x}{x+2}\right)<0$, then $x$ can be

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

Answer: (B), (C)

$0$

For $\log$ to be defined,
$\frac{3-x}{x+2}>0$
$\Rightarrow x\in (-2,3)\cdots \left(1\right)$

Now, ${\log }_{0.5}\left(\frac{3-x}{x+2}\right)<0$
$\Rightarrow \frac{3-x}{x+2}>1[\text{For}0<a<1,{\log }_{a}x\text{is decreasing}]$
$\Rightarrow \frac{3-x}{x+2}-1>0$
$\Rightarrow \frac{3-x-x-2}{x+2}>0$
$\Rightarrow \frac{1-2x}{x+2}>0$
$\Rightarrow x\in (-2,\frac{1}{2})\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$-2<x<\frac{1}{2}$