II. f(x) doubles for each increase of 1 in the x values. Thus, r must be 2, and so we our ar^1 = 6 from ( I ) above becomes f(x) = a*2^x. Applying the restriction ar^1 = 6 results in f(1) = a*2^1 = 6, or a = 3.
Then f(x) = ar^x becomes f(x) = 3*2^2 (Answer A)