Algorithms - Solving recurrence-relations/Bounds?
First of all, a recurrence is not necessarily about the running time of anything. So you don't figure out "the running time", you solve the recurrence.

Second, your recurrence only possibly makes sense for powers of 2, and even then, it needs a base case. I'll assume that $T(1) = C$.

You can use the master theorem to solve your recurrence, but you can also do it directly. First you expand it (for simplicity, I'm assuming the base of the logarithm is 2):
$$
\begin{align*}
T(2^k) &amp;= k + 2T(2^{k-1}) \\ &amp;=
k + 2(k-1) + 4T(2^{k-2}) \\ &amp;=
k + 2(k-1) + 4(k-2) + \cdots + 2^{k-1}(k-(k-1)) + 2^k T(1) \\ &amp;=
(1+2+\cdots+2^{k-1})+(1+2+\cdots+2^{k-2})+\cdots+(1) + 2^k C \\ &amp;=
(2^k-1)+(2^{k-1}-1)+\cdots+(2-1)+(1-1) + 2^kC \\ &amp;=
(2^k+\cdots+1)-(k+1) + 2^kC \\ &amp;=
2^{k+1}-k-2 + 2^kC.
\end{align*}
$$
If you want to be very pedantic, you can prove this by induction. In terms of $n=2^k$, this reads
$$
T(n) = (2+C)n-\log n - 2 \sim (2+C) n.
$$
In particular, $T(n) = \Theta(n)$. You could also get this conclusion from case 1 of the master theorem.