Prove or disprove ${(log_{4} n )}^{2} \in \Theta(log_{2} (n^4))$ I do not know how to start here, thought it would be a good idea to convert them to the same base? ${(log_{4} n )}^{2} = {\left(\frac{ln (n)}{ln (4)}\right)}^{2}$ and ${log_{2}(n^4)}$ = $\frac{ln(n^4)}{ln(2)}$
However I do not know whether this approach is correct and what to do with what I have now.
$$(\log_{4} n) \in \Theta ( \log_2 n^4)$$ so intuition is that statement is wrong.$(\log_4 n)^2$ grows much faster than $\log_2 n^4$.Now I will prove that $\log_2 n^4 \in \textbf{o} (\log_4 n)^2$ which proves that your proposition is wrong.
$$\lim_{n \rightarrow \infty} \frac{\log_2 n^4 }{ (\log_4 n)^2} = 0$$
first note that $|\frac{\log_2 n^4}{(\log_4 n)^2}| = |\frac{4.\log_2 n}{ (\log_4 n) . (\log_4 n)}| \leq C \frac{1}{\log_4n}$ hence we have that limit is $0$ and your proposition is wrong