How Math.Pow (and so on) actually works

Question

So I was googling for a long time and i found almost nothing. I found some info about possible implementation of Math.Pow from this url, but they are inaccurate, for example

Answer 1

pow is usually evaluated by this formula:

x^y = exp2(y*log2(x))

Functions exp2(x),log2(x) are directly implemented in FPU. If you want to implement bignums then they can also be evaluated by basic operators with use of precomputed table of sqrt-powers like:

2^1/2, 2^1/4, 2^1/8, 2^1/16, 2^1/32 ...

to speed up the process