$f(n) = \log n$
$a = 2, b = 2 \implies n^{\log_b a} = n$
So, $f(n) = \log n = O\left(n^{1-\epsilon} \right)$, we can take any $\epsilon$ from $0$-$1$ for example $0.5$ which gives $\log n = O(\sqrt(n))$, proof of which is given here: http://math.stackexchange.com/questions/145739/prove-that-logn-o-sqrtn
So, Master theorem Case 1, and answer will be $O\left(n^{\log_2 2}\right) = O(n)$
Alternate way:
$T(1) = 1 \\ T(2) = 2T(1) + \log 2 = 3 = 3n - 2\\T(4) = 2T(2) + \log 4 = 8 = 3n - 4 \\T(8) = 2T(4) + \log 8 = 19 = 3n - 5\\ T(16) = 2T(8) + \log 16 = 42 = 3n - 6$
The second term being subtracted is growing at a lower rate than the first term. So, we can say $T(n) = O(n)$.
Correct Answer: $A$