We must first note that #4096 = 2^12#. We are not yet sure how x will fit into the matter
If each increase of #x# by 1 drops our #y# value by a factor of 2, it appears that we are dividing by a factor of #2^x#. To check this, we can examine each #x# and #y# value. We will use #f(x)# to denote our potential function, and we will use #y(x)# to denote our known y values.
#x=0, y(0) = 4096, f(0) = 2^12/2^0 = 4096/1 = 4096#
#x=1, y(1) = 2048, f(1) = 2^12/2^1 = 2^11 = 2048#
#x=2, y(2) = 1024, f(2) = 2^12/2^2 = 2^10 = 1024#
#x=3, y(3) = 512, f(3) = 2^12/2^3=2^9= 512#
#x=4, y(4) = 256, f(4) = 2^12/2^4 = 2^8 = 256#
This exponential function models the data accurately. Thus, we can confidently say #y(x) = 2^(12-x) = 2^12/2^x#
