Solution of Logarithmic Inequalities: $\log_{0.5}(\log_{5}(x^2-4))>\log_{0.5}1$ 
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 $(-3,-2)\cup(2,3)$

From the inequality we get 
$x^2-4>1$
Or$x^2>5$
Which provide one condition.
