Let $\log_p$ denote the $p$-adic logarithm and $\mathbb Z_p$ be the ring of integers.
We know that $p$-adic logarithm has no base, but still we have $\log_p(x^n)=n \log_p(x)$ for $n \in \mathbb Z$ and $x \in 1+p \mathbb Z_p$. Here the prime $p$ doesn't mean base. Can we extend the result from $\mathbb Z$ to $\mathbb Z_p$ ? i.e.,
Is $\log_p(x^z)=z \log_p(x)$ for $z \in \mathbb Z_p$ ?
Let $ \mathcal{C}(\mathbb Z_p, \mathbb Q_p)$ denotes the continuous $p$-adic valued functions, then $\log_p \in \mathcal{C}(\mathbb Z_p, \mathbb Q_p)$. Then \begin{align} &\mathbb Z \to \mathcal{C}(\mathbb Z_p, \mathbb Q_p) \\ & n \mapsto \log_p(x^n)=n \log_p(x), \end{align} which is continuous map with respect to $p$-adic topology. So I think the maps extends to $$z \mapsto z\log_p(x),$$ in other words, $\log_p(x^z)=z\log_p(x)$.
I appreciate comments.