Does every basic exponent property have a corresponding logarithm property and vice versa?

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Here is what I'm referring to:

For $0 < a \neq 1$, $\,0 < b \neq 1$, and $m,n \in \mathbb{R^+}$, $p = a^m$, $q = a^n,$ $$\begin{align*} a^0 &= 1 \qquad &\qquad \log_a(1) &= 0 \tag{a}\\[1ex] a^1 &= a \qquad &\qquad \log_a(a) &=1 \tag{b}\\[1ex] a^m a^n &= a^{m + n} \qquad &\qquad \log_a(pq) &= \log_a(p) + \log_a(q) \tag{c}\\[1ex] \frac{a^m}{a^n} &= a^{m - n} \qquad &\qquad \log_a\left(\frac{p}{q}\right) &= \log_a(p) - \log_a(q) \tag{d}\\[1ex] (a^m)^n &= a^{mn} \qquad &\qquad \log_a(p^n) &= n\log_a(p) \tag{e}\\[1ex] \text{(Eq. 1)}& \qquad &\qquad \log_{a^m}(q) &= \frac{1}{m} \log_a(q) \tag{f}\\[1ex] \text{(Eq. 2)}& \qquad &\qquad \log_a(p) &= \frac{\log_b(p)}{\log_b(a)} \Leftrightarrow \log_b(p) = \log_b(a) \log_a(p) \tag{g}\\[1ex] a^m b^m &= (ab)^m \qquad &\qquad \text{(Eq. 3)}& \tag{h} \end{align*}$$

My question is, do equations $1\unicode{x2013}3$ exist? I've been trying to find the right corresponding expressions but to no avail. Thank you.