How do units work out in logarithms?

Question

In a problem I am doing, it boiled down to an integral that resulted in $$\ln(x+1\text{ m})\Big|_{x=0\text{ m}}^{x=3\text{ m}}$$ this is basically just $${\color{red}{\ln(4\text{ m}) - \ln(1\text{ m})}} = {\color{blue}{\ln\left(\frac{4\text{ m}}{1\text{ m}}\right)}} = \ln(4)$$

Obviously, when you you do use logarithm properties to combine the logarithms (blue), the units cancel out and we can a unitless value.

However, if we don't (red), how do the units work in this case? We'd have units of $\ln(\text{meters})$, which doesn't seem to make sense as this is supposedly equivalent to a unitless value.

What am I missing?

Answer 1

A quantity with a unit contains a scalar and the unit as separate quantities $$\underbrace{4}_{\text{scalar}}\cdot \underbrace{\text{m}}_{\text{unit}}$$

Remember your logarithm properties that $\ln(x\cdot y) = \ln(x)+\ln(y)$, so we have $$\ln(4\text{ m}) - \ln(1\text{ m}) = \ln(4) + \ln(\text m) - \ln(1)-\ln(\text m) = \ln(4)-\ln(1) = \ln(4)$$

In this case the units actually cancel, so both expressions are the same.

Answer 2

There are problems where respecting the dimensions is omitted because it entails additional effort for writing things down, but doesn't change the result. (Or are people just working sloppily? )

For example when calculating the potential of a point charge in two dimensions, from the fundamental solution of the Laplacian operator you get $$ \phi(r) = - \frac{q}{2 \pi \varepsilon_0} \ln\frac{r}{r_0} $$ where $r_0$ is a length constant the value of which is arbitrary. Once you're interested in physical quantities based on potential differences you'll see that the value of $r_0$ doesn't matter. (And so sometimes you'll see people setting, for simplicity, $\ln r_0 = 0$ and thus $\ln\frac{r}{r_0} = \ln r - \ln r_0 = \ln r$, which, as you rightly realized, is dangerous... Setting $\ln r_0 = 0$ implies working with dimensionless lengths.)

Much the same, in your problem if you want to be cautious, you could write $$ \ln\left(\frac{x + a}{r_0}\right) $$ ($a = 1\,$m) instead of what you are now using. $r_0$ fixes the dimension and you'll see that its value is irrelevant to the result. Perhaps you could trace back to an earlier stage in your calculation to find where you'd start inserting this additional constant... ;)

Answer 3

I believe that it comes from the first formula, the interior of a logarithm must necessarily be dimensionless. It should be divided by a unit of length, that is, $\ln{\frac{``whatever"}{1 ~ m}}$

Answer 4

You can look at your sum in the following way.

If an object is 3 metres long then that means is that the object is three times longer than the unit of length $(1\,\rm m)$, ie the object is $3\times (1\,\rm m)$ long.

Put another way the object is $\dfrac {3\times (1\,\rm m)}{(1\,\rm m)} = 3$ (a number) times as long as a metre.

$x= 3 \,\rm m$ is an abbreviated form of $3\times (1\,\rm m)$ which can be written as $x/m = 3$ because 3 is the numerical value of when expressed in the unit $\rm m$ and is simply a number.

Thus $\ln(x+1\text{ m})\Big|_{x=0\text{ m}}^{x=3\text{ m}}$ could be written as $\ln((x+1)/\text{ m}))\Big|_{x/\text{m}=0}^{x/\text{m}=3}$ but I think that $\ln((x+1)/\text{ m}))\Big|_{x=0}^{x=3}$ will do.