Question

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

• A. -2 < x < 0
• B. x ∈ (-2, 3)
• C. -2 < x <
• D. -2 < x < 3

Answer 1

The correct option is C

-2 < x <

For log to be defined

${\log }_{0.5}$ $\frac{3-x}{x+2}$ < 0

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

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

x lies -2 < x < 3 ------------(1)

${\log }_{0.5}$ $\frac{3-x}{x+2}$ < 0 ------------(2)

Base of log lies between 0 to 1,then inequality is equivalent to

$\frac{3-x}{x+2}$ > $(0.5{)}^{\circ }$

$\frac{3-x}{x+2}$ > 1
$\frac{1-2x}{x+2}$ > 0
x lies -2 < x < \frac{1}{2} --------(3)
from (1),(2),(3), -2 < x < $\frac{1}{2}$