Question

The set of real values of $x$ for which ${\log }_{0.5}\frac{3-x}{x+2}<0$ is

Answer 1

Given logarithmic ineqaulity as:
${\log }_{0.5}\frac{3-x}{x+2}<0$

Now, for logarithm to be defined, we have

$\frac{3-x}{x+2}>0$

$\Rightarrow \frac{x-3}{x+2}<0$

Solving, we get: $-2<x<3\cdots \left(1\right)$

Since, the base of logarithm lies between 0 to 1, thus the inequality is equivalent to
$\Rightarrow \frac{3-x}{x+2}>(0.5{)}^0$

$\Rightarrow \frac{3-x}{x+2}>1$

$\Rightarrow \frac{3-x-x-2}{x+2}>0$

$\Rightarrow \frac{1-2x}{x+2}>0$

$\Rightarrow \frac{2x-1}{x+2}<0$

Now, Using wavy curve method



$\Rightarrow \frac{2x-1}{x+2}<0\text{for}x\in (-2,\frac{1}{2})\cdots \left(2\right)$
Thus, from $\left(1\right)\&\left(2\right),$ we get the solution set as:
$x\in (-2,\frac{1}{2})$
Hence the correct answer is Option A.