Solving Equation with Euler's Number

Question

How to Approximate "$n$" in

$$1-e^\frac{-n^2}{2N} = \frac{1}{2}?$$

Textbook Answer:

Answer 1

$$1-e^\frac{-n^2}{2N} \approx \frac{1}{2}\iff-e^\frac{-n^2}{2N} \approx \frac{1}{2}-1\iff e^\frac{-n^2}{2N} \approx \frac{1}{2}\iff\frac{-n^2}{2N} \approx\log \frac12\\\iff n^2 \approx2N\log 2\iff n \approx\sqrt{2\log 2}\sqrt N$$

Answer 2

First make it so that the part with the exponential is alone like $$-e^{-\frac{n^2}{2N}}\approx -\frac{1}{2}$$Then take logs of both sides getting $$\frac{n^2}{2N}=\log\bigg(-\frac{1}{2}\bigg)$$ Using log laws you can see how the last line is achieved.

Answer 3

$$1-\exp\left(-\frac{n^2}{2N}\right) \approx\frac12$$

Hence

$$\exp\left(-\frac{n^2}{2N}\right) \approx\frac12$$

Taking logarithm,

$$-\frac{n^2}{2N} \approx \log\left(\frac12 \right)=-\log 2$$

Multiplying by $-2N$,

$$n^2 \approx 2N\log 2 $$

Taking square root would give you the desired result.