Solving $T(n) = 2T(n/2) + T(n-1)/\log n$
By comparison to the recurrence $T(n) = 2T(n/2)$, we see that $T(n) = \Omega(n)$. By comparison to the recurrence $T(n) = 2T(n/2) + n$, we see that actually $T(n) = \Omega(n\log n)$.

Now let us prove by induction that the other direction holds as well. I will interpret $T(n/2)$ as $T(\lfloor n/2 \rfloor)$. Suppose that $T(m) \leq Cm\log m$ for all $m &lt; n$. Then
$$
T(n) \leq 2C\frac{n}{2}\log \frac{n}{2} + \frac{n\log n}{\log n} = Cn\log n + (1-C)n. 
$$
For $C \geq 1$, it follows that $T(n) \leq Cn\log n$.

It seems that we have proved that $T(n) = O(n\log n)$, but actually the base case of the induction doesn't hold. However, this shouldn't be too hard to fix (left to the reader).