import collections
import math
import timeit

def power_bit_length(x):
    return 2**(x-1).bit_length()

def shift_bit_length(x):
    return 1<<(x-1).bit_length()

def power_log(x):
    return 2**(math.ceil(math.log(x, 2)))

def test(f):
    collections.deque((f(i) for i in range(1, 1000001)), maxlen=0)

def timetest(f):
    print('{}: {}'.format(timeit.timeit(lambda: test(f), number=10),
                          f.__name__))

timetest(power_bit_length)
timetest(shift_bit_length)
timetest(power_log)

4.38817000389: power_bit_length
3.69475698471: shift_bit_length
7.91623902321: power_log

6.566169916652143: power_bit_length
3.098236607853323: shift_bit_length
9.982460380066186: power_log

2.8580930233: power_bit_length
2.49524712563: shift_bit_length
3.4371240139: power_log

def next_greater_power_of_2(x):
    return 2**(x-1).bit_length()

In [144]: math.log(6,2)
Out[144]: 2.584962500721156

In [145]: math.log(8,2)
Out[145]: 3.0

In [146]: math.ceil(math.log(6,2))
Out[146]: 3.0

In [147]: math.ceil(math.log(8,2))
Out[147]: 3.0

In [148]: math.ceil(math.log(16,2))
Out[148]: 4.0

In [149]: 2**math.ceil(math.log(6,2))
Out[149]: 8.0

v+=(v==0);
v--;
v|=v>>1;
v|=v>>2;
v|=v>>4;
v|=v>>8;
v|=v>>16;
v++;

def next_power_of_2(n):
    if n == 0:
        return 1
    if n & (n - 1) == 0:
        return n
    while n & (n - 1) > 0:
        n &= (n - 1)
    return n << 1