# Does the logarithm of a non-dimensionless quantity make any sense?

In equation (7) you have the expression $$−P\ln(P−Rv(n+1))+P\ln(P)$$

But since this a difference between two logarithms you can rewrite the expression (remember $\ln a - \ln b = \ln \frac ab$) as $$P\ln\left(\frac{P}{P−Rv(n+1)}\right)$$

Now you have the logarithm of a dimension-less quantity, as it should be.

The reason behind this problem is that you've not yet simplified the final expression.

For example, let's suppose you get a term $\ln (f(v))$ in your final indefinite integral, where $f(v)$ *has dimensions* and isn't dimensionless. This is, as you noted, weird as a logarithm's arguments should **always be dimensionless**. But, now if you apply the limits, you get

$$\ln(f(v))\biggr|_{v_1}^{v_2} = \boxed{\ln\left(\frac{f(v_1)}{f(v_2)}\right)}$$

Now, as you see, the boxed expression is perfectly valid. The argument in the logarithm is, as expected, dimensionless. So, there will never be a case where you'd encounter an expression like $\ln(\text{quantity with dimension})$ if you apply the limits and then analyze the expression.

$$I(v)=e^{\int h(v)dv}=e^{-\ln(P-Rv(n+1))}=\frac{1}{P-Rv(n+1)} \tag{6}$$

If $v$ is has non-trivial dimensions, then $\int \frac{1}{v} dv = \ln\left|\frac{v}{D}\right|$, where $D$ is equivalent to $e^{-C}$ in the dimensionless case: $\int \frac{1}{x} dx = \ln\left|x\right| + C = \ln\left|x\right| - \ln e^{-C}$.

If $v$ is in $\left.\mathrm{m}\middle/\mathrm{s}\right.$, for example, we could write:

$$\int \frac{1}{v} dv = \ln\left|\frac{v}{1\ \left.\mathrm{m}\middle/\mathrm{s}\right.}\right| + C$$