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So my recursive equation is T(n) = 2T(n/2) + log n

I used the master theorem and I find that a = 2, b =2 and d = 1.

which is case 2. So the solution should be O(n^1 log n) which is O(n log n)

I looked online and some found it O(n). I'm confused

Can anyone tell me how it's not O(n log n) ?

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Mater theorem requires that $f(n) = n^c, c\in Z$. In your example, you cannot apply the master theorem directly. Here is my explanation for the answer. We always have $n \geq \log n$. Thus, $T(\frac{n}{2}) + \log n \leq T(\frac{n}{2}) + n$. We can apply master theorem to $T(\frac{n}{2}) + n$ which is $O(n \log n)$. This is a valid bound but not the tightest bound. If you need the tightest bound, I suggest the tree recursion method but algebra can be messy.

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