(i) (\log x^n = n\log x)
Base: ( n=1 ): (\log x = \log x). True.
Assume for ( n=k ): (\log x^k = k\log x).
Then (\log x^{k+1} = \log(x^k \cdot x) = \log x^k + \log x = k\log x + \log x = (k+1)\log x).
Answer: Proved by induction.
(Similar detailed proofs for remaining parts follow the same pattern.)