Assume that $\mathsf{NP} \subseteq \mathsf{P}/\text{log(n)}$, does it imply that $\mathsf{P} = \mathsf{NP}$? I am trying to either prove or refute the claim mentioned in the title.
Any ideas ?
Yes. All advice strings of length log n can be cycled though in polynomial time.
The polynomial time algorithm for SAT would be:
For each of the polynomially many advice strings, try to use it to produce a satisfying assignment for SAT (using self-reducibility) in polynomial time. I would guess this is in the original Karp Lipton paper.