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Q. The set of real values of x for which $\log_{0.2} \frac{x +2}{x} \leq 1$ is

Linear Inequalities

Solution:

The inequality is $log_{0.2} \frac{x + 2}{x} \leq 1$
The L.H.S is valid if $ \frac{x + 2}{x} > 0$
$ \Rightarrow \: x(x + 2) > 0 \Rightarrow x < -2 \, or \: x > 0$
Solving the inequality, we get (note that base < 1).
$ \frac{x + 2}{x}\geq 0.2 = \frac{1}{5}$
$\Rightarrow \frac{x +2}{x} -\frac{1}{5} \ge 0 \Rightarrow \frac{4x + 10}{5x} \ge 0$
$x \left(2x+5\right)\ge0 \Rightarrow x \le -\frac{5}{2}$ or $x \ge0$
Taking the intersection, we get
$x \leq - \frac{5}{2}$ or $x > 0$
$\Rightarrow \: x \in \left( - \infty , - \frac{5}{2} \right] \cup (0 , \infty)$