Question

The solution set of ${\log }_2|4-5x|>2$ is equal to

• A. $\left(\frac{8}{5},+\infty \right)$
• B. $\left(\frac{4}{5},\frac{8}{5}\right)$
• C. $(-\infty ,0)\cup \left(\frac{8}{5},+\infty \right)$
• D. none of these

Answer 1

The correct option is C $(-\infty ,0)\cup \left(\frac{8}{5},+\infty \right)$

${\log }_2|4-5x|>2\Rightarrow |4-5x|>4$
Case 1:
$4-5x>4\Rightarrow x<0\therefore x\in (-\infty ,0)$
Case 2:
$-4+5x>4\Rightarrow x>\frac{8}{5}\therefore x\in (\frac{8}{5},\infty )$
From Case 1 & Case 2;
$x\in (-\infty ,0)\cup (\frac{8}{5},\infty )$

Ans: C