Question

The set of values of $x$ for which inequality holds true
${\log }_{0.2}|x-3|\ge 0$

Answer 1

Given:
${\log }_{0.2}|x-3|\ge 0$

$\text{Base of the logarithm lies between}0\text{to}1,$
then inequality is equivalent to

$\Rightarrow |x-3|\le (0.2{)}^0\text{and}|x-3|\ne 0$

$\Rightarrow |x-3|\le 1\text{and}|x-3|\ne 0$

$\Rightarrow -1\le x-3\le 1\text{and}|x-3|\ne 0$

$\Rightarrow -1+3\le x\le 1+3\text{and}|x-3|\ne 0$

$\Rightarrow 2\le x\le 4\text{and}|x-3|\ne 0$

$\therefore x\in [2,4]-\left\{3\right\}$

Hence the correct answer is Option B.