How should I compute log to the base two in python. Eg. I have this equation where I am using log base 2

import math
e = -(t/T)* math.log((t/T)[, 2])

It’s good to know that

but also know that math.log takes an optional second argument which allows you to specify the base:

In [22]: import math

In [23]: math.log?
Type:       builtin_function_or_method
Base Class: <type 'builtin_function_or_method'>
String Form:    <built-in function log>
Namespace:  Interactive
Docstring:
    log(x[, base]) -> the logarithm of x to the given base.
    If the base not specified, returns the natural logarithm (base e) of x.


In [25]: math.log(8,2)
Out[25]: 3.0

float → float math.log2(x)

import math

log2 = math.log(x, 2.0)
log2 = math.log2(x)   # python 3.3 or later

• Thanks @akashchandrakar and @unutbu.

float → int math.frexp(x)

If all you need is the integer part of log base 2 of a floating point number, extracting the exponent is pretty efficient:

log2int_slow = int(math.floor(math.log(x, 2.0)))
log2int_fast = math.frexp(x)[1] - 1

• Python frexp() calls the C function frexp() which just grabs and tweaks the exponent.

• Python frexp() returns a tuple (mantissa, exponent). So [1] gets the exponent part.

• For integral powers of 2 the exponent is one more than you might expect. For example 32 is stored as 0.5×2⁶. This explains the - 1 above. Also works for 1/32 which is stored as 0.5×2⁻⁴.

• Floors toward negative infinity, so log₂31 computed this way is 4 not 5. log₂(1/17) is -5 not -4.

int → int x.bit_length()

If both input and output are integers, this native integer method could be very efficient:

log2int_faster = x.bit_length() - 1

• - 1 because 2ⁿ requires n+1 bits. Works for very large integers, e.g. 2**10000.

• Floors toward negative infinity, so log₂31 computed this way is 4 not 5.

If you are on python 3.3 or above then it already has a built-in function for computing log2(x)

import math
'finds log base2 of x'
answer = math.log2(x)

If you are on older version of python then you can do like this

import math
'finds log base2 of x'
answer = math.log(x)/math.log(2)

Using numpy:

In [1]: import numpy as np

In [2]: np.log2?
Type:           function
Base Class:     <type 'function'>
String Form:    <function log2 at 0x03049030>
Namespace:      Interactive
File:           c:\python26\lib\site-packages\numpy\lib\ufunclike.py
Definition:     np.log2(x, y=None)
Docstring:
    Return the base 2 logarithm of the input array, element-wise.

Parameters
----------
x : array_like
  Input array.
y : array_like
  Optional output array with the same shape as `x`.

Returns
-------
y : ndarray
  The logarithm to the base 2 of `x` element-wise.
  NaNs are returned where `x` is negative.

See Also
--------
log, log1p, log10

Examples
--------
>>> np.log2([-1, 2, 4])
array([ NaN,   1.,   2.])

In [3]: np.log2(8)
Out[3]: 3.0

http://en.wikipedia.org/wiki/Binary_logarithm

def lg(x, tol=1e-13):
  res = 0.0

  # Integer part
  while x<1:
    res -= 1
    x *= 2
  while x>=2:
    res += 1
    x /= 2

  # Fractional part
  fp = 1.0
  while fp>=tol:
    fp /= 2
    x *= x
    if x >= 2:
        x /= 2
        res += fp

  return res

>>> def log2( x ):
...     return math.log( x ) / math.log( 2 )
... 
>>> log2( 2 )
1.0
>>> log2( 4 )
2.0
>>> log2( 8 )
3.0
>>> log2( 2.4 )
1.2630344058337937
>>> 

Try this ,

import math
print(math.log(8,2))  # math.log(number,base) 

logbase2(x) = log(x)/log(2)

In python 3 or above, math class has the fallowing functions

import math

math.log2(x)
math.log10(x)
math.log1p(x)

or you can generally use math.log(x, base) for any base you want.

log_base_2(x) = log(x) / log(2)

Don’t forget that log[base A] x = log[base B] x / log[base B] A.

So if you only have log (for natural log) and log10 (for base-10 log), you can use

myLog2Answer = log10(myInput) / log10(2)