I've been told it is never seen in physics, and "bad taste" to have it in cases of being the argument of a logarithmic function or the function raised to $e$. I can't seem to understand why, although I suppose it would be weird to raise a dimensionless number to the power of something with a dimension.
Answer
It's not "bad taste", it's uncalculable to the point of meaninglessness.
The whole point of dimensional analysis is that there are some quantities that are not comparable to each other: you can't decide whether one meter is bigger or smaller than ten amperes, and trying to add five volts to ten kelvin will only yield inoperable nonsense. (For details on why, see What justifies dimensional analysis? and its many linked duplicates on the sidebar on the right.)
This is precisely what goes on with, say, the exponential function: if you wanted the exponential of one meter, then you'd need to be able to make sense of $$ \exp(1\:\rm m) = 1 + (1\:\rm m) + \frac12(1\:\rm m)^2 + \frac{1}{3!}(1\:\rm m)^3 + \cdots, $$ and that requires you to be able to add and compare lengths with areas, volumes, and other powers of position. You can try to just trim out the units and deal with it, but keep in mind that it needs to match, exactly, the equivalent $$ \exp(100\:\rm cm) = 1 + (100\:\rm cm) + \frac12(100\:\rm cm)^2 + \frac{1}{3!}(100\:\rm cm)^3 + \cdots, $$ and there's just no invariant way to do it.
Now, to be clear, the issue is much deeper than that: the real problem with $\exp(1\:\rm m)$ is that there's simply no meaningful way to define it a way that will (i) be independent of the system of units, and (ii) keep a set of properties that will really earn it the name of an exponential. If what one wants is a simple clear-cut way to see it, a good angle is noting that, if one were to define $\exp(x)$ for $x$ with nontrivial dimension, then among other things you'd ask it to obey the property $$ \frac{\mathrm d}{\mathrm dx}\exp(x)=\exp(x), $$ which is dimensionally inconsistent if $x$ (and therefore $\mathrm d/\mathrm dx$) is not dimensionless.
It's also been noted in the comments, and indeed in a published paper, that you can indeed have Taylor series over dimensional quantities, by simply setting $f(x) = \sum_{n=0}^\infty \frac{1}{n!} \frac{\mathrm d^nf}{\mathrm dx^n}(0)x^n$, and that's true enough. However, for the transcendental functions we don't want any old Taylor series, we want the canonical ones: they're often the definition of the functions to begin with, and if someone were to propose a definition of, say $\sin(x)$ for dimensionful $x$, then unless it can link back to the canonical Taylor series, it's simply not worth the name. And, as explained above, the canonical Taylor series have fundamental scaling problems that render them dead in the water.
That said, for logarithms you can on certain very specific occasions talk about the logarithm of a dimensional quantity $q$, but there you're essentially taking some representative $q_0$ and calculating $$\log(q/q_0)=\log(q)-\log(q_0),$$ where in making sense of the latter you require that the two numerical values be in the same units ─ in which case the final answer is independent of the unit itself. If the situation also allows you to drop additive constants, or incorporate them into something else (such as when solving ODEs, for example, with a representative case being the electrostatic potential of an infinite line charge, or when doing plots in log scale) then you might get rid of the $\log(q_0)$ in the understanding that it will come out in the wash when you come back to dot the i's.
However, just because it can be done in the specific case of the logarithm, which is unique in turning multiplicative constants into additive ones, doesn't mean you can use it in other contexts ─ and you can't.
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