Exponential or logarithm of a dimensionful quantity?

The only sensible rule when working with units is, that you can only add together terms which carry the same unit. Say $ [x]=[y] $, then $x+y$ is unit-wise a valid statement. You may also multiply arbitrary units together. Whether that is physically sensible is another question. Obviously you cannot add, e.g meters and seconds, but multiplying to form $m/s$ as a unit for velocity is a valid operation.

From that follows, that the argument of the exponential must not carry a unit, because the exponential is defined as a power series. $$ e^x =\sum_{n=0}^\infty \frac{x^n}{n!}$$ If $x$ were to carry a unit, say meters, one would add (schematically) $m+m^2+m^3+\cdots$, which is nonsenical.

If you encounter an exponential, a sine/cosine, logarithm,... in physics you will find almost always that its argument, which must be dimensionless, is a product of often two conjugate variables. Examples are time and frequency, or distance and momentum.

See "what's the logarithm of a kilometer" for a discussion about that. As David Z also said in the comment here, using the logarithm of a dimensionful quantity is actually quite reasonable.

This is not true for the exponential. The power series definition "proves" that, however the same argument would also work for the logarithm. Personally I don't like treating the Taylor series as anything more than a useful calculation tool. The "more fundamental" (of course there's no such metric) definition is as a solution to the differential equation $\tfrac{\mathrm{d}\exp}{\mathrm{d}x} = \exp(x)$. Which tells you right away $$ \tfrac{[\exp]}{[x]} = [\exp] \qquad \Rightarrow\quad [x] = 1. $$ Note that this does not come out when using the analogous definition of the logarithm: $$ \frac{\mathrm{d} \ln}{\mathrm{d}x} = \frac{1}{x} \qquad \Rightarrow \quad \tfrac{[\ln]}{[x]} = \tfrac{1}{[x]} \qquad \Rightarrow \quad [x] =\:? $$ Of course, both equations only define the functions up to gauge of an initial value. For $\ln(1) = 0$ to make sense, you certainly need the argument to be dimensionless. But as long as you only consider differences between logarithms, the gauge cancels anyway!