Let $\log_a$ be the D866: Standard real logarithm function in base $a \in (0, \infty) \setminus \{ 1 \}$.
Let $e$ be the D169: Napier's constant.
Let $e$ be the D169: Napier's constant.
Then
\begin{equation}
\log_a e
= \frac{1}{\log_e a}
\end{equation}