Consider this initial-rate data at a certain temperature for the reaction described by 2no2

NO2 ... O3 ... rate

0.65 ... 0.8 ... 2.6x10^4

1.1 ... 0.8 ... 4.4x10^4

1.76 ... 1.4 ... 12.32x10^4

rate2/rate1 = k[NO2 2]^a[O3 2]^b / k[NO2 1]^a[O3 1]^b

K and O3 cancel

4.4 / 2.6 = (1.1 / 0.65) ^a

1.7 = 1.7^a ... a = 1

rate 3 / rate 2 = k[NO2 3][O3 3]^b / k[NO2 2][O3 2]^b

k cancels

since you know the order of the rxn with respect to NO2, plug in the values and solve the equation. the [O3]^b is a separate value so reduce all numbers to a final value and then solve for b

12.32 / 4.4 = (1.76) (1.4) ^b / (1.1) (0.8) ^b ... 1.76/1.1 = 1.6 ... 1.4 / 0.8 = 1.75

2.8 = 1.6 x (1.75) ^b

1.75 = (1.75) ^b ... b = 1

rate = k[NO2][O3]

2.6x10^4 / (0.65) (0.8) = k

k = 5x10^4