I have some integer count suppose '51' and I want represent that much of integers in binary number. Here I need to do log(51) so i get some float value as 3.93182563272. But I want it in some integer format as 4 which could be used to represent 51 integers.
Log value can be calculated as
import math
math.log(51)
If you want the number of binary digits, that would be base 2, whereas math.log by default returns the natural logarithm (base e). A second argument can be used to specify an alternative base. Then you can use math.ceil to round up the number.
math.ceil(math.log(51, 2))
6.0
You haven't specified a python version but if you have python 3, (thanks @delnan), you can use math.log2 instead, which should be more accurate:
math.ceil(math.log2(51))
6.0
numpy also has a log2 method (but is probably overkill for this application).
math.ceil actually returns a float, so if you want an integer you can wrap the expression in int:
int(math.ceil(math.log(51, 2)))
6
By the way, there is also the function bin which you might want to look at. It returns a string containing the binary representation of an integer:
bin(51)
'0b110011'
...but if you don't want to mess around with any of that (thanks again @delnan), you can just use bit_length instead:
(51).bit_length()
6