Logaritm confusion: How is $\ln((N/2)^2-x^2)=\ln((N/2)^2)+\ln(1-(2x/N)^2)$?

Question

I'm doing a thermal physics problem and in the notes they do, for $x<<N$ this:

$$\ln\left(\left(\frac{N}{2}\right)^2-x^2\right)=\ln\left(\left(\frac{N}{2}\right)^2\right)+\ln\left(1-\left(\frac{2x}{N}\right)^2\right).\tag1$$

This is not obvious to me. First I thought they simply went like

$$\ln\left(\left(\frac{N}{2}\right)^2-x^2\right)=\ln\left(\left(\frac{N}{2}+x\right)\left(\frac{N}{2}-x\right)\right)=\ln\left(\frac{N}{2}+x\right) + \ln\left(\frac{N}{2}-x\right),$$

but this is not at all on the same form as $(1)$. Can someone explain the equation?

Answer 1

Observe that, by factoring out the left term, and then simplifying a bit,

$$\begin{align} \left( \frac{N}{2} \right)^2 - x^2 &= \left( \frac{N}{2} \right)^2 \cdot \left( 1 - \frac{x^2}{(N/2)^2} \right) \\ &= \left( \frac{N}{2} \right)^2 \cdot \left( 1 - \frac{4x^2}{N^2} \right) \\ &= \left( \frac{N}{2} \right)^2 \cdot \left( 1 - \left( \frac{2x}{N} \right)^2 \right) \end{align}$$

From here, you can use the property $\ln(ab) = \ln(a) + \ln(b)$ to get your desired result.