If y=cbx then we must use a system of equations with the given points.
Equation 1: using (1,54) we have 54=cb
Equation 2: using (2,16) we have 16=cb-2=c(1/b2)
Solving for c in equation 1 we have (54/b)=c
We plug this into the second equation to obtain: 16=(54/b)(1/b2)=54/b3
Solving for b3 we get b3=54/16=27/8 yielding b=3/2.
Using equation 1 we get 54=c(3/2) and solving for c we get c=36
Thus: y=36(3/2)x