Let the rate law be expressed as: $\text{Rate} = k[A]^x[B]^y$

According to the first condition, when the concentration of $B$ is doubled, the rate is doubled: $2 \times \text{Rate} = k[A]^x(2[B])^y$ Dividing this by the initial rate equation gives: $2 = 2^y \implies y = 1$

According to the second condition, when the concentrations of both $A$ and $B$ are doubled, the rate increases by a factor of 8: $8 \times \text{Rate} = k(2[A])^x(2[B])^y$ Dividing this by the initial rate equation gives: $8 = 2^x \times 2^y$ Substituting the value of $y = 1$: $8 = 2^x \times 2^1 \implies 2^x = 4 \implies x = 2$

Therefore, the overall rate law for the reaction is $\text{Rate} = k[A]^2[B]^1 = k[A]^2[B]$.