Prove that the number of words in a positive integer a in base Beta is 1 + ilog(a)/BPW where Beta = pow(2, BPW)

Question

The number of digits (words) in a base Beta of any positive integer a is 1 + floor(log(a)) where the log is in base Beta. When the base Beta = 2 this is just the number of bits. That is 1 + lg(a) is the number of bits. I want to prove that floor(log(a)) = floor(lg(a))/BPW where BPW is the number of bits per word and lg is the log in base 2. So if Beta = pow(2, BPW) then I want to prove what always seemed obvious; the number of words in base Beta is 1 + floor(lg(a)/BPW) = 1 + floor(lg(a))/BPW but I am struggling with the floor function. How can I be sure that floor(lg(a)/BPW) = floor(lg(a))/BPW?