$\sigma>0$, $a>1$ and $C$ are constant real number. Does the following holds for any $x>\sigma$?
$\log[ ( x+\sigma)^a - (x-\sigma)^a ] = (a-1) \log(x) + C$.
Background
John Tukey, on the book Exploratory Data Analysis (1977, chapter 4), proposes a method to transform sample data from different populations so that the spread of the transformed data is independent of the median of the samples. He suggest that the transformation, which he calls "re-expression", helps the description and comparison of the populations. This webpage summarizes the method.
The book follows a hands-on approach and gives almost no mathematical justification for the techniques it proposes.
I got curious about the "re-expression" and, by writing down what is described on the book, found that it would work if the following equation is satisfied for any $M$, given $\sigma$, $a$ and $C$ independent of $M$.
$\log[ ( M+\sigma)^a - (M-\sigma)^a ] = (a-1) \log(M) + C$.
I checked it numerically, and this expression seems to be true, or at least a very good approximation. Could you please help me derive it algebraically?
This is equivalent to saying that $$ e^C = \frac{(x+\sigma)^a-(x-\sigma)^a}{x^{a-1}}. $$ This is not true, but if we simplify the right-hand side, we find $$ \frac{(x+\sigma)^a-(x-\sigma)^a}{x^{a-1}} = x \left( \left(1+\frac{\sigma}{x}\right)^a - \left(1-\frac{\sigma}{x}\right)^a \right), $$ and expanding this as a power series in $1/x$ gives $$ x \left( \left(1+\frac{\sigma}{x}\right)^a - \left(1-\frac{\sigma}{x}\right)^a \right) = 0 + 2a \sigma + 0 + \frac{a(a-1)(a-2)}{3} \frac{\sigma^3}{x^2} + O(x^{-4}), $$ or taking the log again, $$ \log{((x+\sigma)^a-(x-\sigma)^a)} = (a-1)\log{x} + \log{(2a\sigma)} + \frac{(a-1)(a-2)}{6}\frac{\sigma^2}{x^2}+O(x^{-4}). $$ Presumably in applications the latter terms are sufficiently small that neglecting them does not influence the data significantly.