Kevin Ford in his slides on prime gaps explains how the simple idea that primes occurs with probability $1/\log(x)$ can be used to construct a probabilistic model for prime gaps.
source: https://conf.math.illinois.edu/~ford/montreal_talk1_primegaps.pdf
On slide 6 (reproduced below) he explains that the probability of a prime gap of size $g$ is
$$\left(1-\frac{1}{\log(x)} \right)^g$$
This makes sense if we assume the probability of a number being prime is independent of another.
He then says this is
$$\approx e^{g/\log N}$$
Question: I don't understand how this is N and not $k$.
I'd welcome answers that are suitable for readers not trained to university maths.
