Solution of Logarithmic Inequalities: $\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1$

Question

If $\log_{0.5}(\log_5 (x^2-4)) >\log_{0.5}1$ then x lies in the interval:
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3)$
(a) $(-3,-\sqrt{5})\cup(\sqrt{5}, 3\sqrt{5})$
(c) $(\sqrt{5}, 3\sqrt{5})$
(d) $\phi $

I have solved quite a few logarithmic inequalities and I am familiar with the procedure. I was just not able to get the answer that matches with the given options in the question.

My Working:
$\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1 $
$\log_{5}(x^2-4)<1$
$\log_{5}(x^2-4)<\log_5(5)$
$x^2-4<5$
$x^2<9$

$\therefore -3<x<3$

Now the second equation comes from the fact that
$x^2-4>0$
$\therefore x<-2, x>2$

The intervals formed are $(-3,-2)\cup(2,3)$

Answer 1

First of all, $\log_{0.5}(\log_5 (x^2-4))$ will remain defined iff

$\log_5 (x^2-4)>0\iff x^2-4>5^0=1\iff x^2>5\iff$ either $x>\sqrt5$ or $x<-\sqrt5\ \ \ \ (1)$

$$\log_{0.5}(\log_5(x^2-4))>\log_{0.5}1\iff\dfrac{\log_e(\log(x^2-5))}{\log_e(0.5)}>\dfrac{\log1}{\log_e(0.5)}=0$$

Now as $0.5=\dfrac12=2^{-1},\log_e(0.5)=\log_e(2^{-1})=-\log_e2$

$\implies0>\dfrac{\log_e(\log_5(x^2-4))}{\log_e(0.5)}=-\dfrac{\log_e(\log_5(x^2-4))}{\log_e2}$

$\implies\log_e(\log_5(x^2-4))<0\iff\log_5(x^2-4)<e^0=1$

$\iff x^2-4<5^1=5\iff x^2<5+4\iff-3<x<3 \ \ \ \ (2)$

What is the intersection of $(1),(2)?$

Answer 2

From the inequality we get

$x^2-4>1$

Or$x^2>5$

Which provide one condition.