About the space needed for segment tree, i came across the following equation

N = (ceil)((log(N) / log(2)) + 1)
size = 1 << N

interesting part is , the formula used natural logarithms, i.e, base ‘e’. I also read that upper limit of space required for segment tree is (4 * size of array)

The base e is added by us for convenience. We know that ln(a)/ln(b) is log(a) with base (b)…Thus, we can easily remove the base e and work with base 2 if we want to…

Important to note, \text{log}_2(N)=\text{log}_e(N)/\text{log}_e(2) so (\text{ceil})((\text{log}_e(N) / \text{log}_e(2)) + 1) = (\text{ceil})(\text{log}_2(N)+1). So what you have there has nothing to do with the natural logarithm, it is just written in a strange way.