Exponential or logarithm of a dimensionful quantity? I have a unit measure, say, seconds, $s$. Furthermore let's say I have a dimensionful quantity $r$ that is measure in seconds, $s$. What is the unit measure of $e^r$? ($1/r$ is in $Hz$.)
My question is general, how to find the unit measure of a transformation function $y=f(x)$ where $x$ takes some known unit measure. I give above two functions $f(\cdot)=e^\cdot$ and $f(.)=1/\cdot$.
 A: 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.
A: 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!
A: Consider:
Conc = 100 mg/mL
Log10( Conc ) = 2
What if I express Conc as mg/dL?  Then Conc2 = 1 mg/dL (note that this is the same, just different units with a different quantity):
Log10( Conc2 ) = 0.
Log10( Conc ) ne Log10( Conc2 ) and we have an issue, unless we retain units.  Why wouldn't log10 mg/mL be any less reasonable than mg/mL?
Second, consider this argument from the internet: to take the logarithm, the quantity must be dimensionless, thus divide by the unit (under what justification?):
Log10( 10 km / 1 km )
The issue is:
Log10( 10 km / 1 km ) = Log10( 10 km ) - Log10( 1 km )
-Kevin
