the base is inside the ()

y=2log(3)x

y+1=log(3)9x

what i did:

y=2log(3)x

y=log(3)9x-1

2log(3)x = log(3)9x-1

2log(3)x = 2log(3)3X - log(3)3

2log(3)x - 2log(3)3x = - log(3)3

x(2log(3)1 - 2log(3)3) = -log(3)3

x = -log(3)3 / (2log(3)1 - 2log(3)3)

x = -1 / (0-2)

x = 1/2

however, the answer is x=3

could you tell me hwere i went wrong please?