# 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 $f(.)=1/\cdot$.

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Unless $r(s)$ is unitless, $e^r$ doesn't make much sense (see, for example, its definition in terms of the power series) – Kyle Kanos Apr 25 '14 at 15:42
@Jika: As Kyle mentioned, in physics it is impossible to have something of the form $e^r$, unless $r$ is unitless. If obtain something of the form $e^r$ where $r$ is not unitless, that means you made a mistake somewhere. Try to check your math for errors. – DumpsterDoofus Apr 25 '14 at 15:46

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 $x$ where 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.

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People do sometimes write things like $\log E$ and so on, with $E$ being a quantity with units, but that's understood to be a shorthand for $\log\frac{E}{E_0}$ for some reference value $E_0$ whose value is irrelevant. (And as far as I'm concerned, it's generally better to write out $\log\frac{E}{E_0}$ explicitly.) – David Z Apr 25 '14 at 16:42
I would say always, not almost always, although it might not be readily apparent. You might have $(e^r)^k$, where $k$ has units of inverse meters. With logs you might have $(\ln{r} - \ln{k})$. An author might choose to drop the second term if it is of no consequence, but that only serves to add to the confusion. – garyp Apr 25 '14 at 16:44
Another common notation (especially in axis labels) is something like $\log(d/\mathrm{m})$ (in this case the $E_0$ David Z mentions is just "one metre"), or $\log(v/c)$ (the "$E_0$" is a dimensional constant with the same dimensions as the quantity of interest). – Kyle Oman Apr 25 '14 at 20:13

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 "prooves" 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!

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