Given that, $\log_{2}\left[3+\log_{3}\{4+\log_{4}(x-1)\}\right]-2=0$
$\Rightarrow \log_{2}\left[3+\log_{3}\{4+\log_{4}(x-1)\}\right]=2$
$\Rightarrow 3+\log_{3}\{4+\log_{4}(x-1)\}=2^{2}\quad \left[\because \log_{a}{b} = x \Rightarrow b=a^{x}\right]$
$\Rightarrow \log_{3}\{4+\log_{4}(x-1)\}=1$
$\Rightarrow 4+\log_{4}(x-1)=3^{1}$
$\Rightarrow \log_{4}(x-1)=-1$
$\Rightarrow x-1=4^{-1}$
$\Rightarrow x= \frac{1}{4}+1$
$\Rightarrow \boxed{x= \frac{5}{4}}$
$\therefore$ The value of $4x = 4 \times \frac{5}{4} = 5.$
Correct Answer $:5$