Better way to compute floor of log(n,b) for integers n and b?

Question

I'm looking to compute floor(log(n,b)) where n and b are both integers. Directly implementing this function fails for even slightly large values of n and b

# direct implementation
def floor_log(n,b):
    return math.floor(math.log(n,b))

For example, floor_log(100**3, 100) evaluates to 2 instead of the correct value 3.

I was able to come up with a working function which repeatedly divides until nothing remains

# loop based implementation
def floor_log(n,b):
    val = 0
    n = n // b
    while n > 0:
        val += 1
        n = n // b
    return val

is there a faster or more elegant way of obtaining this solution? Perhaps using built-in functionality?

Answer 1

One way is to make your function self-correcting:

def floor_log(n,b):
    res = math.floor(math.log(n,b))
    return res + 1 if b**(res+1) == n else res 

Or a bit more obscure:

def floor_log(n,b):
    res = math.floor(math.log(n,b))
    return res + (b**(res+1) == n)