Explanation
$$\textbf{Step 1 : The principle properties of logarithm:} \log_am^n = n\log_am \left( \textbf{Where m,a>0 and a}\neq 1\right)$$
Consider,
$${{\log }_{4}}\dfrac{{{t}^{2}}}{4}-2{{\log }_{4}}4{{t}^{4}}$$
Put $$t=-2$$ in above expression, We get
$$ ={{\log }_{4}}\dfrac{{{\left( -2 \right)}^{2}}}{4}-2{{\log }_{4}}4{{\left( -2 \right)}^{4}} $$
$$ ={{\log }_{4}}\left( \dfrac{4}{4}\right)-2{{\log }_{4}}\left(4\times 16\right) $$
$$ ={{\log }_{4}}1-2{{\log }_{4}}64 $$
$$ =0-2{{\log }_{4}}{{4}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\log }_{a}}1=0 \right) $$
$$ =-2\times 3{{\log }_{4}}4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m \right) $$
$$ =-6\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \therefore {{\log }_{a}}a=1 \right) $$
Hence,$$-6$$ is the answer.
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