To prove or disprove mathematically $n \log_{3}{n^2} \in \Theta(n(\log_3 n)^2) $

Question

I have to prove mathematically that

$$n\log_{3}{n^2} \in \Theta(n(\log_3 n)^2) $$

and

$$4^n\log_3n \in o(6^n)?$$

So anyone can at least give me a hint where to start? Which proof method should I use?

Answer 1

For the first statement (considering the $O$-notation, since, as hinted in comments, $\Theta$-result is impossible) you may merely transform $n*log_3n^2$ to $2*n*log_3n$. Then you should note that $log_3n$ grows slower than $(log_3n)^2$ and that it is possible to find a constant $M$ for a given $n_0$, s.t. $2*n*log_3n\leq M*n*(log_3n)^2$ for all $n_0\leq n$.

For the second statement, note that $6^n = 4^n*(3/2)^n$, and again $(3/2)^n$ grows much faster than $log_3n$ and to finish the proof explicitly you should present some number $n_0$ for a given $\epsilon$, s.t. $4^n*log_3n\leq \epsilon\ 6^n$ for all $n_0\leq n$.