Natural log approximation I've got an equation that contains 
$$x^p - 1$$ 
$x$ is any positive number (such as 2) and $p$ is a small positive number close to 0 (such as 0.001).
For some reason (that I may have known in High School), this can be approximated by 
$$\ln(x) \times p$$
Can anyone explain why this approximation works?  Does it have something to do with approximating a natural log with a series?
 A: I have nothing to add to whuber's brilliant answer using the Box-Cox transformation.  I just wanted to offer an alternative source of the approximation, using the Maclaurin series for the natural logarithm:
$$\ln (y+1) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots.$$
Ignoring the higher-order terms in the expansion gives the crude first-order approximation:
$$\ln (y+1) \approx y.$$
Substituting $y=x^p-1$ gives the approximation in your question:
$$p \ln(x) = \ln(x^p) = \ln(y+1) \approx y = x^p-1.$$
Thus, in addition to being derivable from the Box-Cox transform, this approximation can be seen as a crude first-order Maclaurin approximation to the logarithm (which is indeed an approximation you might have run across in High School).
A: The map $$x \to \frac{x^p-1}{p}$$ is known in statistics as a Box-Cox transformation.  The power $p$ may be any real number except $0$, for which the formula at the right is undefined (it would divide zero by zero).  However, for any fixed positive $x$ the limit as $p$ approaches $0$ can readily be found by L'Hopital's Rule as
$$\lim_{p\to 0} \frac{x^p-1}{p} = \lim_{p\to 0} \frac{\frac{d}{dp}(x^p-1)}{\frac{d}{dp}(p)} = \lim_{p\to 0} \frac{\log(x) \exp(p\log x)}{1} = \log(x) $$
where I computed
$$\frac{d}{dp} x^p = \frac{d}{dp} \exp(p\log x) = \log(x) \exp(p\log x)$$
using the Chain Rule and exploited the continuity of $\exp$ to find the limit simply by setting $p=0$ in the right hand side.
This establishes $p=0$ as a removable singularity in the Box-Cox family and helps us understand the "zero power" as naturally corresponding to the logarithm.

Edit
Alternatively, recognize that this transformation can also be expressed as
$$x \to \int_1^x y^{p-1}\mathrm{d}y.\tag{*}$$
Simply set $p=0$ to obtain
$$x \to \int_1^x y^{-1}\mathrm{d}y = \log(x)$$
(because that's the definition of the natural logarithm).
So, to answer the question, notice that the function $p \to y^{p-1} = \exp((p-1)\log(y))$ is continuous because $\exp$ is continuous. Therefore the integral $(*)$ is a continuous (even a differentiable) function of $p,$ whence when $p\approx 0,$ the transformation will be approximated by $p=0:$ the logarithm.  The closeness of the approximation, though, varies with $y,$ getting worse as $y\to 0$ or $y\to \infty.$
