Math support ilog() optimisation: count digits when base is a power of 2

[saxbophone]

• $log_2(x)$ is $countbits(x)..countbits(x) + 1$
  where $countbits()$ just counts the number of bits to the right of the highest bit set, including the highest bit set.
• $log_{BASE}(x)$ is $digitscount(x)..digitscount(x) + 1$
• $log()$ for other base $b$ that is a power of 2 may be calculated quickly by taking $countbits(x)$ and dividing it by $countbits(b)$

Note: For the first two versions, $+1$ is only added to the ceiling part of the calculation if any bits to the right of the most significant set bit are set. Otherwise, $floor() = ceil()$ (because $x$ is an integer power of the base: $x = b^n$).

Logarithms are very useful for underpinning other mathematical operations so this optimisation may be very useful.

[saxbophone]

Found useful optimisations to be applied here when checking if a single digit is a power of 2 and for log2(): http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogObvious

This optimisation should also probably be applied to Nat.bit_length()

[saxbophone]

Extension task:

• [ ] Use optimisation for $log_2$ to slightly increase the speed of $log_b$ by using the $log_b(x) = log(x) / log(b)$ relation to gain a starting $[floor, ceil]$ estimate that is closer to the answer, to then iterate from using binary search, rather than gradually building up to the answer using repeated divisions. May not be faster or may even be slower for small values, but may give a good performance increase for log of much larger values...