In a reaction, $$A + B \to $$  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

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