log(A/B) = log(A)-log(B), where A and B have units

Question

We all know that

$$\log\left(\frac{A}{B}\right) = \log(A)-\log(B)$$

However, in the case where $A$ and $B$ have (identical) units, such as kilograms, the right-hand side cannot be performed because the arguments are not dimensionless. How can one rationalize this observation without making the circular argument to convert the right-hand side back to the left-hand side when you have units?

Answer 1

Treat the unit like a variable:

$$\log\left(\dfrac{A \cdot kg}{B \cdot kg}\right) = \log(A) + \log(kg) - \log(B) - \log(kg) = \log(A) - \log(B)$$

After all, that's what you're doing when you say the units cancel.

Answer 2

In the physical sciences we often come across quantities (like potentials, phases, and levels) such that only differences (or quotients) of quantities are directly observable, but we still want a workable calculus of such quantities without having to choose a reference value. This idea is made rigorous by the notion of torsor—a group that has "forgotten" its identity.

In your case

• the domain of $\log$ is the "multiplicative" torsor of values that a positive quantity can take (where we have forgotten what units the quantity is given in)
• the range is the "additive" torsor of levels for that quantity (where we have forgotten the reference level), and
• the equality given is the statement that $\log$ is a homomorphism between these two torsors. If you had a definition of $\log$ that didn't depend on the $\log$ between the underlying groups, you could potentially prove that this is a homomorphism directly.