5 x 10^7 M-1s-1, k-1 = 6 x 10^5s-1 and k2 = 1.

I know the answer for the question which is 41 mM, but i dont understand how they got this please please explain

For a Michaelis enzyme, k1 = 1.5 x 10^7 M-1s-1, k-1 = 6 x 10^5s-1 and k2 = 1.5 x 10^4s-1. What is the substrate concentration at half Vmax