Integer root of floating point value

[RonShepard]

@mecej4

A good working solution for the cube root problem would probably require a good way of generating sufficiently close first approximations without resorting to calling exp or log.

When one does these things by hand, the first step is to divide the exponent by 3 for a cube root (2 for a square root, etc). If more accuracy is needed, then extracting maybe the high-order 4 bits of the mantissa and using that in a lookup table would be sufficient.