Hello.<span><span> </span><span> <span><span> ------- Let 2003 be the zero year; then 2005 is the three year, and 2008 the 5 year. --------------- P = ab^x --- P(3) = ab^3 = 800000 P(0) = ab^0 = 900000
--- a = 900000 Solve for "b":: b^3 = 8/9 b = 2/cbrt(9) ---- Equation:: P(x) = 900000^x ---- Ans: P(5) = 900000