ln ⁡ ( x ) ≈ a x 1 / a − a , which...

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
$a{x}^{\frac{1}{a}}-a=\log x+O\left(\frac{(\log x{)}^2}{a}\right)$
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 $a{x}^{\frac{1}{a}}-a$! It's not as if $a{x}^{\frac{1}{a}}-a$ is particularly easy to compute for large $a$.

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
$\begin{array}{rl}\underset{a\to 0^{+}}{lim}\frac{x^a-1}{a}& =\underset{a\to 0^{+}}{lim}\frac{e^{a\log x}-1}{a}\\ & =\log (x);\end{array}$
in other words, the derivative at $0$ of $a\mapsto x^a$ is $\log x$.