I noticed there was a double precision divide after all log2 calls when using cygwin64 so I wrote a small program to illustrate it. Both loops give the same result for the 20 decimal places that are printed:

#include 
#include 
#include 

int main() {
  uint32_t i;
  for (i = 1 ; i <= 10 ; i++)
    printf("log %u: %.20llf\n", i, log2((double)i)); 
  for (i = 1 ; i <= 10 ; i++)
    printf("log %u: %.20llf\n", i, 1.442695040888963387 * log((double)i));
}

Looking at the relevant parts of the assembly code shows this for the log2 calculation for the first loop, which apparently uses natural logs and converts to log2:

.L2:
    pxor    %xmm0, %xmm0
    cvtsi2sdl   %ebx, %xmm0
    call    log
    movl    %ebx, %edx
    movq    %rsi, %rcx
    addl    $1, %ebx
    divsd   %xmm6, %xmm0
    movq    %xmm0, %r8
    movapd  %xmm0, %xmm2
    call    printf
    cmpl    $11, %ebx
    jne .L2

And this for the log2 calculation for the second loop:

.L3:
    pxor    %xmm0, %xmm0
    cvtsi2sdl   %ebx, %xmm0
    call    log
    movl    %ebx, %edx
    movq    %rsi, %rcx
    addl    $1, %ebx
    mulsd   %xmm6, %xmm0
    movq    %xmm0, %r8
    movapd  %xmm0, %xmm2
    call    printf
    cmpl    $11, %ebx
    jne .L3

The question is: Why does cygwin64 divide by log2(e) instead of multiply by 1 / log2(e)? Aren't double precision multiplies a lot faster than double precision divides? Would the use of a divide be specific to cygwin64 or would all versions of gcc use divide instead of multiply?