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As we know, the integral of $\frac{1}{x}$ is $ln(x)+c$. Because $x$ and $dx$ have the same dimension, $\int\frac{dx}{x}$ is dimensionless. But my problem is: $x$ is dimensional. I've been trained that the natural log of a dimensional quantity is meaningless, and yet here it is.

Furthermore, it seems like this crops up a lot, such as in separable ODE's. E.g. let's say we have $dx/dt=k x$, where $x$ is dimensional, we obtain $ln(x)=rt+c$. Both sides are dimensionless. But then we take the exponential of both sides and get $x=k e^{rt}$ -- suddenly both sides have dimensions, and $k$ has magically changed from $e^c$, which is dimensionless, into dimensional $k$, but where did the dimension come from? It seems like every textbook I've read is sloppy about this.

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  • $\begingroup$ do you mean the notation is sloppy? The integral is a sum of all the infinitesimal bits of size dx/x, no? $\endgroup$ – Llouis May 10 at 15:26
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$\ln{(x)} + C = \ln{(Cx)}$. The dimensions of $C$ are those which are needed to make $Cx$ dimensionless. :)

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  • $\begingroup$ Here we can even talk about the logarithm of a dimension and unit, e.g. $[x] = \log(L),$ $[C] = \log(L^{-1}) = -\log(L).$ $\endgroup$ – md2perpe May 10 at 16:01
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Both members of $$\frac{\dot x}x=r$$ are in $s^{-1}$ (speed over distance).

After integration

$$\int\frac{dx}x=\log\left(\frac x{x_0}\right)=r(t-t_0),$$

both members are dimensionless, which allows you to take the exponential:

$$\frac x{x_0}=e^{r(t-t_0)}.$$

Of course,

$$x=x_0e^{r(t-t_0)}$$ are in $m$.

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