This should work:

from math import log10
def rev(num):
    if num < 10:
        return num
    else:
        ones = num % 10
        rest = num // 10
        #print ones, rest, int(log10(rest) + 1), ones * 10 ** int(log10(rest) + 1)
        return ones * 10 ** int(log10(rest) + 1) + rev(rest)
print rev(9000), rev(1234), rev(1234567890123456789)

You could also reduce the number of times you call log10 and number of math operations by using a nested recursive function:

def rev(num):
    def rec(num, tens):
        if num < 10:
            return num        
        else:
            return num % 10 * tens + rec(num // 10, tens // 10)
    return rec(num, 10 ** int(log10(num)))

I'm not going to give you the answer, but I'll give some hints. It looks like you don't want to convert it to a string -- this makes it a more interesting problem, but will result in some funky behavior. For example, reverseDisplay(100) = 1.

However, if you don't yet have a good handle on recursion, I would strongly recommend that you convert the input to a string and try to recursively reverse that string. Once you understand how to do that, an arithmetic approach will be much more straightforward.

Your base case is solid. A digit reversed is that same digit.

def reverseDisplay(n):
    if n < 10:
        return n
    last_digit = # ??? 12345 -> 4
    other_digits = # ??? You'll use last_digit for this. 12345 -> 1234
    return last_digit * 10 ** ??? + reverseDisplay(???)
    # ** is the exponent operator. If the last digit is 5, this is going to be 500...
    # how many zeroes do we want? why?

If you don't want to use any string operations whatsoever, you might have to write your own function for getting the number of digits in an integer. Why? Where will you use it?

Imagine that you have a string 12345.

reverseDisplay(12345) is really 
    5 + reverseDisplay(1234) ->
        4 + reverseDisplay(123) ->
            3 + reverseDisplay(12) ->
                2 + reverseDisplay(1) ->
                    1