Logarithm approximation

Question

I saw on https://math.stackexchange.com/a/2382470/432488 the following approximation: $\ln(x) \approx a x^{1/a}-a$, which is good for large value of $a$. Where does it come from?

Answer 1

For large $a$ we have

$$x^{ \frac{1}{a} } = e^{\frac{\log x}{a}} = 1 + \frac{\log x}{a} + O \left( \left( \frac{\log x}{a} \right)^2 \right)$$

(by Taylor's theorem) which gives

$$ax^{\frac{1}{a}} - a = \log x + O \left( \frac{(\log x)^2}{a} \right)$$

so the approximation is accurate once $a$ is large relative to $(\log x)^2$. By the way, I would describe this in exactly the opposite way: that $\log x$ is a good approximation to $ax^{ \frac{1}{a} } - a$! It's not as if $ax^{ \frac{1}{a} } - a$ is particularly easy to compute for large $a$.

Answer 2

This is the same thing as asserting that, if $a>0$ and $a$ is very small, then$$\log(x)\approx\frac{x^a-1}a,$$which is true, since\begin{align}\lim_{a\to0^+}\frac{x^a-1}a&=\lim_{a\to0^+}\frac{e^{a\log x}-1}a\\&=\log(x);\end{align}in other words, the derivative at $0$ of $a\mapsto x^a$ is $\log x$.