Find value of $k$ such that domain of $$f(x)=\log_{0.5}\left(\log_{4}\left(\log_3[(x-k)^2]\right)\right)$$ is $(-\infty \: 2] \cup [6 \:\infty)$ where $[.]$ is Greatest integer function.
For outside $\log$ to take valid arguments we have $$ 0 \lt \log_{4}\left(\log_3[(x-k)^2]\right)\lt 1$$ which implies
$$1 \lt \log_3[(x-k)^2] \lt 4$$ $\implies$
$$3 \lt [(x-k)^2] \lt 81$$ $\implies$
$$(x-k)^2 \ge 4$$ and $$(x-k)^2 \le 81$$
From here how can we find $k$