For the reaction, $$A + B \to $$  products, it is observed that
(i) On doubling the initial concentration of $$A$$ only, the rate of reaction is also doubled and
(ii) On doubling the initial concentrations of both $$A$$ and $$B,$$ there is a change by a factor of 8 in the rate of the reaction.
The rate of this reaction is, given by

Solution :
$$\eqalign{ & {\text{For the reaction,}} \cr & A + B \to {\text{Products}} \cr} $$
On doubling the initial concentration of $$A$$  only, the rate of the reaction is also doubled, therefore
$$\eqalign{ & {\text{Rate}} \propto {\left[ A \right]^1}\,\,\,...{\text{(i)}} \cr & {\text{Let initial rate law is}} \cr & {\text{Rate}} = k\left[ A \right]{\left[ B \right]^y}\,\,\,...{\text{(ii)}} \cr} $$
If concentration of $$A$$  and $$B$$  both are doubled, the rate gets changed by a factor of 8.
$$\eqalign{ & 8 \times {\text{rate}} = k\left[ {2A} \right]{\left[ {2B} \right]^y}\,\,\,...{\text{(iii)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,{\text{Rate}} \propto {{\left[ A \right]}^1}} \right] \cr & {\text{Dividing Eq}}{\text{. (iii) by Eq}}{\text{. (ii), we get}} \cr & \,\,\,\,\,\,\,8 = 2 \times {2^y} \cr & \,\,\,\,\,\,\,4 = {2^y} \cr & {\left( 2 \right)^2} = {\left( 2 \right)^y} \cr & \therefore \,\,y = 2 \cr & {\text{Hence, rate law is, rate}} = k\left[ A \right]{\left[ B \right]^2} \cr} $$