Find domain of $\log_4(\log_5(\log_3(18x-x^2-77)))$ 
Problem:$\log_4(\log_5(\log_3(18x-x^2-77)))$
Solution:
$\log_3(18x-x^2-77)$ is defined for $(18x-x^2-77) \ge 3$
$(x^2-18x+77) \le -3$
$(x^2-18x+80) < 0$ {As it can't be 0}
$(x-8)(x-10)<0$
$8<x<10$
So domain is $(8,10)$

Am I doing right ??

HINT:
For for real $m>0$ $\log_ma,$ is defined if $a>0$ as if $\log_ma=b\iff a=m^b>0$
$\log_4(\log_5(\log_3(18x-x^2-77)))$ is defined if $\log_5(\log_3(18x-x^2-77))>0$
$\implies \log_3(18x-x^2-77)>5^0=1$
$\implies (18x-x^2-77)>3^1=3$
$\implies x^2-18x+80<0$
Now, the roots of $x^2-18x+80=0$ are $\frac{18\pm\sqrt{18^2-4\cdot1\cdot 80}}2=8,10$
Now, if $(x-a)(x-b)<0$ where $a,b(<a)$ are real 
Either $x-a<0$ and $x-b>0\implies b<x<a$
or  $x-a>0\implies x>a$ and $x-b<0\implies x<b$ which needs $a<x<b$ which is impossible as $a>b$ 
So, the domain here will be $(8,10)$
