How is this approximation related?

Question

I'm in the dark about the following approximation.

Given that $r >> (l_1 + l_2)$

$\phi(P) \approx \dfrac{\lambda}{4\pi\epsilon_0}ln(\dfrac{r+l_2}{r-l_1}) \approx \dfrac{\lambda(l_1 + l_2)}{4\pi\epsilon_0 r}$

I do not see the relation. If possible can someone explain to me this relation?

Answer 1

Hint

$$\frac{r+l_2}{r-l_1}=\frac{r-l_1+l_1+l_2}{r-l_1}=1+\frac{l_1+l_2}{r-l_1}\approx 1+\frac{l_1+l_2}{r}$$ Now, use, for small $x$, $\log(1+x)\approx x$.

Answer 2

$$\log\frac{r+l_1}{r-l_2}=\log\frac{1+(l_1/r)}{1-l_2/r}\\=\log(1+l_1/r)-\log(1-l_2/r)\\ \approx l_1/r-(-l_2/r) = (l_1+l_2)/r$$