What is the dimension of $e^x$ given the dimension of x is L?
My first thought was that it’s a completely different type of unit as $L^3$ is well-defined, but $3^L$ isn’t.
However, I realized that $e^x= \sum_{n=0}^\infty \frac{x^n}{n!}$ has a power series, which would mean that it has all the units all at once.
But the problem is that this power series was formulated with no regard for dimensions, and the original would have been $\sum_{n=0}^\infty \frac{x^n}{n!} \frac{d^n f}{dx^n}$ where the units of $x^n$ in the numerator and $dx^n$ in the denominator cancel out and leave the units of $f$. Now we’re at square one.
There’s one important property of $e^x$ that could be leveraged: $e^a e^b = e^{a+b}$. This means that it has units of $L^0$, $L^\infty$, or $L^0+L^\infty$.
I don’t really know where to go from here.