Problem: To prove $$2n^2 2^n +n\log n=\Theta(n^2 2^n)$$
I initially started by trying to prove the upper bound, and the way I thought for doing this was by principle of mathematical induction, considering $$4n^2 2^n \geq 2n^2 2^n + n\log n.$$ However this proof gets fairly complicated. Here is my approach:
Upper Bound: Let $c=3$, then: $$3n^2 2^n =2n^2 2^n + n^2 2^n \geq 2n^2 2^n + n\log n$$ $$n^2 2^n \geq n\log n$$ $$n 2^n \geq \log n.$$
The last one is true since: $$n \geq \log n ~\forall n\geq 1.$$
Similarly I have proved for the lower bound with $c=1$. My question is, is this proof sufficient or would I have to prove $n>\log n$ for $n>1$ as well? And if yes do I do that using derivatives?
Also, I have a great difficulty in proving $$\frac{6n^3}{\log(n+1)} =\Theta(n^3).$$ Please help me out with this too.
EDIT: The above 2 have been solved.
Last problem: While going about proving $$n^k n^e + n^k logn = \Theta(n^k n^e)$$ for all k and e >0 For proving the upper bound I take the constant k but how can I prove $$kn^e > logn$$ for 1>e>0