Why should transcendental functions and their arguments be dimensionless? While looking for the answer on the internet I came across an answer giving this explanation
 "Another way of seeing clearly why an exponential's argument should be dimensionless is to Taylor expand: $\exp(x)=1+x+x^2/2+\cdots$ Every term has a different dimension if x is dimensionful, and hence cannot be summed."
But this is not a proof in the sense we do in mathematics!
So can anyone provide with a mathematically sound proof?
 A: No proof is needed, it's true simply by definition of units. When dealing with units in any physics calculation, we generally respect the following unofficial axioms that define units:
1) A unit is treated as a variable in any calculation
2) Any number that is linear in a specific unit (with no intercept) is said to be a quantity of "those units".
So if "m" is a unit, that means that we can only classify a quantity as having "units of m" if it is of the form $k \text{m}$ for some unitless number $k$. Respecting the definition above, we see that $x = e^{2 \text{m}} = 1 + 2 \text{m} + 2 \text{m}^2 + \frac{4}{3} \text{m}^3 + \cdots$ is not strictly of form $k \text{m}$, thus by the definition above, it can't be said to have "units of m". 
Aside from that definition, I don't think "units" has a formal definition in pure mathematics (perhaps someone will point me out wrong here), so without that formal definition, you can't make a proof. You need a definition like the one I wrote above.
A: You can think of the equation of motion
$s(t) = s(0) + v(0) t + \frac{a}{2} t^2$
as a Taylor expansion of the function $s(t)$ around $t=0$, exact for constant acceleration. The coefficients of this series are $s(0)$, $v(0)$ and $a$ ($=a(0)$) and have different units.
When you expand a function $f(x)$ in Taylor series, the coefficients are given by
$a_n = \frac{1}{n!} \left[\frac{d^n f(x)}{dx^n}\right]_{x=x_0}$
and so they have units of $f$, $f/x$, $f/x^2$, $f/x^3$, etc.
So, the correct expansion of $exp(x)$ for $x$ in meters is
$exp(x) = 1 m + x + \frac{1}{2m} x^2 + \frac{1}{6m^2} x^3 + ... $
For instance,
$exp(2m) = 1 m + 2m + \frac{1}{2m} 4 m^2 + \frac{1}{6m^2} 8 m^3 + ... $
If you want to keep the expansion as simple as
$exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... $,
you'd better have a dimensionless $x$.
Besides, a unit should be treated as a multiplication factor. Different functions treats multiplied factors different ways. For example,
$log(4.18 J) = log(4.18) + log(J)$ and
$e^{4.18 J} = (e^{J})^{4.18} = (e^{4.18})^{J}$.
That's too complicated and usually avoided in formulas of natural sciences.
In some equations involving entropy, the logarithm is divided as in $log(ab) = log(a) + log(b)$, such that the arguments $a$ and $b$ are dimensional. The Wikipedia page [Sackur-Tetrode equation]1 has an example of such equations, but also a justification for it:

Strictly speaking, the use of dimensioned arguments to the logarithms
is incorrect, however their use is a "shortcut" made for simplicity.
If each logarithmic argument were divided by an unspecified standard
value expressed in terms of an unspecified standard mass, length and
time, these standard values would cancel in the final result, yielding
the same conclusion. The individual entropy terms will not be
absolute, but will rather depend upon the standards chosen, and will
differ with different standards by an additive constant.

