<s>[INST] Why is it that natural log changes are percentage changes? What is about logs that makes this so? Can somebody explain how the properties of logs make it so you can do log linear regressions where the coefficients are interpreted as percentage changes?
 [/INST] These posts all focus on the difference between two values as a proportion of the the first: $\frac{y-x}{x}$ or $\frac{y}{x} - 1$.  They explain why
$$\frac{y}{x} - 1 \approx \ln(\frac{y}{x}) = \ln(y) - \ln(x).$$
You might be interested in the difference as a proportion of the average rather than as a proportion of the first value, $\frac{y-x}{\frac{y+x}{2}}$ rather than $\frac{y-x}{x}$. The difference as a proportion of the average is relevant when comparing methods of measurement, e.g. when doing a Bland-Altman plot with differences proportional to the mean.  In this case, approximating the proportionate difference with the difference in the logarithms is even better:
$$\frac{y-x}{\frac{y+x}{2}} \approx \ln(\frac{y}{x})  .$$
Here is why:
Let $z = \frac{y}{x}$.
$$\frac{y-x}{\frac{y+x}{2}} = \frac{2(z-1)}{z+1}$$
Compare the Taylor series about $z =1$ for $\frac{2(z-1)}{z+1}$ and $\ln(z)$.
$$\frac{2(z-1)}{z+1} = (z-1) - \frac{1}{2}(z-1)^2 + \frac{1}{4}(z-1)^3 + ... + (-1)^{k+1}\frac{1}{2^{k-1}}(z-1)^k + ....$$
$$\ln(z) = (z-1) - \frac{1}{2}(z-1)^2 + \frac{1}{3}(z-1)^3 + ... + (-1)^{k+1}\frac{1}{k}(z-1)^k + ....$$
The series are the same out to the $(z-1)^2$ term.  The approximation works quite well from $z=0.5$ to $z=2$ , i.e., when one of the values is up to twice as large as the other, as shown by this figure.

A value of $z=0.5$ or $z=2$ corresponds to a difference that is 2/3 of the average.
$$\frac{y-x}{\frac{y+x}{2}} = \frac{2x-x}{\frac{2x+x}{2}} = \frac{x}{\frac{3x}{2}} = \frac{2}{3}$$
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