I'm not sure how lower bounds affect the question to the P=NP problem.
Let a SAT instance with a size of n be transformed into an instance of a problem X with a size of n3.
If you find a lower bound of the SAT problem which is $$\Omega(c^n) \ \ \ \ c > 1 $$
What would be the consequence to the problem P=NP?
We already know that SAT has an at exponential lower bound time complexity. But what would be the complexity of the problem X? Will it be $$\Omega((c^n)^3)$$
Will X's complexity affect the problem N=NP in any way? (assuming that the lower bound for SAT exists).
My thought are that since SAT is NP-complete there should be no problem in having an exponential lower bound. Any thoughts?