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.
 A: If $y(x)= \log x$, then $y'(x)=\frac{1}{x}=e^{-y(x)}.$
A: $\ln(x)$ (and every other logarithm) satisfies the linear, homogeneous differential equation
$$
xf''(x) + f'(x) = 0.
$$
For $\ln$, this is admittedly just another way of stating $f'(x) = x^{-1}$.
A: Let $f$ be a bijective function on $I$, with values in $J$, of reciprocal $g$. If $f'$ never cancels, then $g$ has a derivative on $J$, and $g' = \frac{1}{f'(g)}$.
Here $\exp$ is a bijection from $\mathbb R$ to $\mathbb R^*_+$, whose derivative never cancels. Hence, its reciprocal $\log$ is derivable on $\mathbb R_+^*$ and verifies
$$\log'(x) = \frac{1}{\exp'(\log(x))}$$
which is a fancy way of saying that $\log$ verifies the differential equation $y' = \frac 1x$, and the equation $y'=e^{-y}$ as @Fred said.
A: 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 $x'(y) = 1/y$.
A: Since $\ln x:=\int_1^x\frac1t\operatorname dt$, apply FTC.  Get $\ln' x=\frac1x$.
So, $\boxed{f'(x)=\frac 1x}$.
A: You bet it does.
$$y' = \frac{1}{x}$$
Or even better,
$$x \cdot y' = 1$$
.
