Clearly, n log(n) is the dominant term here. If we could prove that n <= n log(n) and log(n) <= n log(n) for all n starting from some n0, then n log(n) + n + log(n) <= 3n log(n) for all n > n0. Note that log(n) >= 1 when n >= 2 (if logarithm is to the base 2).nlogn+n+logn <=M(nlogn)

Hint: In plain text, it is customary to denote exponentiation using ^. For example, n^2 denotes .