(a) Substituting gives	
	110 = A
	174.9 = 110b,
so that			
	b = 174.9/110 = 1.59.
Thus, the model is 	
	C(t) = 110(1.59t).
(b) Putting t = 0.75 gives
	C(0.75) = 110(1.590.75)  155.754.
Thus, the model is accurate to within 0.754 million A.S.M.s per month.
(c) Putting
	400 = 110(1.59t)
gives
	(400/110) = 1.59t
	ln (400/110) = t ln(1.59)
whence
	t = ln(400/110) / ln(1.59))  2.75
That is, in November, 1996.