Question

The solution set of the inequality $(0.5{)}^{{\log }_3{\log }_{0.2}(x^2-\frac{4}{5})}<1$ is

• A. $(-\frac{2}{\sqrt{5}},\frac{2}{\sqrt{5}})$
• B. $(-1,-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},\infty )$
• C. $(-\infty ,-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},1)$
• D. $(-1,-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},1)$

Answer 1

The correct option is D $(-1,-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},1)$

$(0.5{)}^{{\log }_3{\log }_{0.2}(x^2-\frac{4}{5})}<1$
$\log$ function is defined
$\left(i\right)(x^2-\frac{4}{5})>0$
$\Rightarrow x\in (-\infty ,-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},\infty )$

$\left(ii\right){\log }_{0.2}(x^2-\frac{4}{5})>0$
$\Rightarrow (x^2-\frac{4}{5})<1$
$\Rightarrow x\in (-\frac{3}{\sqrt{5}},\frac{3}{\sqrt{5}})$
From $\left(i\right)$ and $\left(ii\right)$
$x\in (-\frac{3}{\sqrt{5}},-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},\frac{3}{\sqrt{5}})...\left(1\right)$

Now, $2^{-{\log }_3{\log }_{0.2}(x^2-\frac{4}{5})}<1$
Taking ${\log }_2$ both sides
$\Rightarrow -{\log }_3{\log }_{0.2}(x^2-\frac{4}{5})<0$
$\Rightarrow {\log }_3{\log }_{0.2}(x^2-\frac{4}{5})>0$
$\Rightarrow {\log }_{\frac{1}{5}}(x^2-\frac{4}{5})>1$
$\Rightarrow (x^2-\frac{4}{5})<\frac{1}{5}$
$\Rightarrow x^2-1<0$
$\Rightarrow x\in (-1,1)...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get
$x\in (-1,-\frac{2}{\sqrt{5}})\cup (\frac{2}{\sqrt{5}},1)$