Final answer:
Both log4 n and log16 n are considered equivalent in the context of big O notation because each can be expressed in terms of the other up to a constant factor, indicating they grow at the same rate.
Step-by-step explanation:
When discussing if log4 n = o(log16n) or if log16n = o(log4n), we're evaluating logarithmic functions within the context of big O notation, which describes the upper bound of a function's growth rate. Big O notation is commonly used in computer science to describe the performance or complexity of an algorithm.
Let's recall a key property of logarithms: changing the base of a logarithm is equivalent to multiplying by a constant. Using this property, we know that:
log₄ n = (log₁₆ n) / (log₁₆ 4)
Since 4 is a power of 16 (4 = 16^(1/2)), the (log₁₆ 4) is a constant. Therefore, log4 n is equal to log16 n up to a constant factor. This means log4 n is O(log16 n) as it grows at the same rate.
Similarly, log16 n can be expressed in terms of base 4, and it will also be just a constant factor different from log4 n, since 16 is a power of 4 (16 = 4^2). Therefore, log16 n is also O(log4 n).
In conclusion, log4 n and log16 n are essentially the same up to constant factors for purposes of big O notation, meaning they grow at the same rate and thus can be considered equivalent in this context.