Is there an approximation to the natural log function at large values?

Almost as Semiclassical answered, write $x=A \times 10^{n-1}$ where $n$ is the number of digits before the decimal point, that is to say $1\leq A < 10$ and use the very fast converging expansion $$\log \left(\frac{1+y}{1-y}\right)=2\sum_{k=0}^{\infty}\frac{y^{2 k+1}}{2 k+1}$$ with $y=\frac{A-1}{A+1}$.

Let us take an example : $x=123456789$; then $A=1.23456789$ and $n=8$; so $y \approx 0.104972$. Now, let us look at the value of $$S_p=2\sum_{k=0}^{p}\frac{y^{2 k+1}}{2 k+1}$$ For the first values of $p$, the sums are successively $0.2099447424$, $0.2107158833$, $0.2107209817$, $0.2107210219$, $0.2107210222$ which is the solution for ten exact decimal places.

So, by the end $$\log^{(p)}(x)=(n-1)\log(10)+ S_p$$ which leads to the successive values of $18.63062549$, $18.63139663$ ,$18.63140173$, $18.63140177$, $18.63140177$ for an exact value equal to $\approx 18.63140177$.

for any positive $x$

$\ln \left( x \right) \approx a{x^{\frac{1}{a}}} - a$
where $a$ is any large constant

The larger $a$, the better approximation.

Hint: $~\ln x=-\ln\dfrac1x~:~$ Can you take it from here ?