Q.Prove the following by using method of induction: $\log_a x^n = n \log_a x$, $x > 0$, $n \in N$.

Let $P(n):\log_ax^n=n\log_ax$. Base: $n=1$: $\log_ax=1\cdot\log_ax$; trivially true. Hypothesis: assume $\log_ax^k=k\log_ax$. Step: $\log_ax^{k+1}=\log_a(x^k\cdot x)=\log_ax^k+\log_ax$ (product rule of logarithms) $=k\log_ax+\log_ax=(k+1)\log_ax$, matching $P(k+1)$. * …

Base case trivial, step uses the log product rule $\log_a(PQ)=\log_aP+\log_aQ$ (a pre-establishe …

Trying to prove the log product rule from scratch instead of citing it as a known alge …