Question

Find the set of real values of x for which ${\log }_{0.2}$ $\frac{x+2}{x}$ $\le$ 1, is ________

• A. U (0,)
• B. U (0,)
• C. U (0,)
• D. U (0,)

Answer 1

The correct option is C

U (0,)

${\log }_{0.2}$ $\frac{x+2}{x}$ $\le$ 1

For log to be defined

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

x > 0 or x < -2 ------------(1)

Given inequality ${\log }_{0.2}$ $\frac{x+2}{x}$ $\le$ 1

Since base of logarithm is 0.2 which lies between 0 to 1, then inequality is equivalent to

$\frac{x+2}{x}$ $\ge$ $(0.2{)}^1$

$\frac{x+2}{x}$ $\ge$ 0.2

$\frac{4x+10}{5x}$ $\ge$ 0
x is in $(-\infty ,\frac{-5}{2}]$ U (0,$\infty$) ___ (2)
From (1) & (2),
$xϵ$ $(-\infty ,\frac{-5}{2}]$ U (0,$\infty$)