Does the logarithm satisfy any differential equation? The exponential function $\exp:\mathbb{R}\to\mathbb{R}_+$ satisfies the differential equation $f^\prime(x)=f(x)$.
Does the logarithm $\log:\mathbb{R}_+\to\mathbb{R}$ also satisfy any differential equation?
Curious about if this gets closed immediately or not... seems like a way too basic question, but I have been totally stuck with this since yesterday.

Here's a fun way to see the relationship between the definition for the exponential function and the "obvious" answer:
Suppose $y = e^x$.  By definition of the exponential function, this means that $dy/dx = y$.  Now view $x(y)$ as a function of $y$, i.e., $x$ is the inverse of the exponential function.  Rearranging the relationship between the differentials above, we have
$$
\frac{dy}{dx} = y \quad \Rightarrow \quad \frac{dx}{dy} = \frac{1}{y}.
$$
Thus, the inverse of the exponential function obeys the differential equation $x'(y) = 1/y$.
