Big-O Notation regarding logarithms

Question

I got asked an interview question that wanted me to discern the Big-O notation of several logarithmic functions. The functions were as follows:

f(x) = log⁵

Answer 1

This is a matter of math:

1. f(x) = log⁵(x)
2. f(x) = log(x⁵) = 5 * log x
3. f(x) = log(6*log x) = log 6 + log(log x)
4. f(x) = log(log x)

So the Big O is

1. O(log⁵(x))
2. O(log x)
3. O(log (log x))
4. O(log (log x))

So (1) and (2) aren't equivalent, but (3) and (4) are (though they're different from both (1) and (2))

Answer 2

f(x) = log^5(n)
f(x) = log(n^5) -> 5 log(n)
O(5 log(n)) < O(log(n)^5)

f(x) = log(6*log n) -> log(6)+log(log(n))
f(x) = log(log n) 
log(log n) < log(6) + log(log(n)) 

, although log(6) is a constant so they have same O

Answer 3

1. log⁵ x is the same as writing log log log log log x, which is a very slow-growing function of x.
2. This is equivalent to 5 log x (rewriting exponentiation inside the log as multiplication outside), which is equivalent to log x.
3. This is equivalent to log 6 + log log x, which is equivalent to log log x.
4. This is just log log x.

So you have O(log log log log log x), O(log x), O(log log x) and O(log log x), three distinct Big-O classes.

If your interviewer said 3 and 4 were different, either he was mistaken or you've misremembered the question (happens all the time).

Answer 4

I'll assume you mean f(n), not f(x). For 1 and 2, log^5(n) is equivalent to O(log log log log log(n)), while log(n^5) = 5 log(n) = O(log n).

For 3 and 4, I disagree. log(6*log n) = log(6) + log(log n) = O(log log n), which is the same as 4 - O(log log n).