In a reaction, $A+B \rightarrow$ Product, rate is doubled when the concentration of $B$ is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants $(A$ and $B$ ) are doubled. Rate law for the reaction can be written as

Let the order of reaction with respect to $A$ and $B$ is $x$ and $y$ respectively. So, the rate law can be given as $R=k[A]^x[B]^y$ When the concentration of only $B$ is doubled, the rate is doubled, so $R_1=k[A]^x[2 B]^y=2 R$ If concentrations of both the reactants $A$ and $B$ are doubled, the rate increases by a factor of 8 , so $\begin{aligned} R^{\prime \prime}=k[2 A]^x[2 B]^y & =8 R \\ \Rightarrow k 2^x 2^y[A]^x[B]^y & =8 R \end{aligned}$ From Eq. (i) and (ii), we get $\begin{array}{rlrl} & \Rightarrow & \frac{2 R}{R} & =\frac{[A]^x[2 B]^y}{[A]^x[B]^y} \\ & \therefore & =2^y \\ & y & =1 \end{array}$ From Eq. (i) and (iv), we get $\Rightarrow \frac{8 R}{R}=\frac{2^x 2^y[A]^x[B]^y}{[A]^x[B]^y}$ or $8=2^x 2^y$ Substitution of the value of $y$ gives, $\begin{aligned} & 8=2^x 2^1 \\ & 4=2^x \end{aligned}$ $\begin{aligned} & & (2)^2 & =(2)^x \\ \therefore & & x & =2 \end{aligned}$ Substitution of the value of $x$ and $y$ in Eq. (i) gives, $R=k[A]^2[B]$