This is my passing 64ms solution in Python without using bitwise operations. One problem I ran into with this approach is that the machine's inaccurate storage of floating-point values led to erroneous results for large values of n, so I rounded the result out to 11 decimal places.

```
import math
class Solution(object):
def isPowerOfTwo(self, n):
"""
:type n: int
:rtype: bool
"""
if n <= 0:
return False
calc = math.log(n)/math.log(2) % 1
calc = round(calc, 11)
return calc == 0.0
```