The population 01 Egypt in 2002 was 74224400 and was 7614140
Solution
Let the initial population be P0 ant in t years it becomes P(t) exponentially with a growth factor r (yearly), then
P(t) = P0 rt ... ... ... (1)
In above problem we have,
P0 = 74,224,400 (The starting year 2002)
After t = 4 (i.e. in 2006) the population becomes P(t) | t=4 = 76,141,400
(a) Using these data in equation (1), (Four year growth rate b, t = 1, (one interval of four years) we get
76,141,400 = 74,224,400 b, or, b = 76,141,400 / 74,224,400
hence, b = 1.0258 (The four year growth factor, taking one interval of 4 years)
(b) The four year percent change = (76,141,400 - 74,224,400) x 100 / 74,224,400
= 2.583 %
(c) One year growth rate r, is achieved using equation (1) with t = 4, we get
76,141,400 = 74,224,400 r4, or, r4 = 76,141,400 / 74,224,400
hence, r = 1.0064 (The one year growth factor, taking four intervals of 1 year)
(d) One year percent change = (r -1) x 100 = 0.0064 x 100 = 0.64 %
(e) The function m that determines the population in terms of number of years is P(t) given in equation (1) with yearly growth rate. Therefore,
P0 rt = m (t) = 74,224,400 (1.0064)t ... .... (2)
(f) Let in t years the population of the Egypt will be doubled, then
m(t) = 2 P0 = P0 (1.0064)t, Therefore, t log (1.0064) = log 2
or, t = log 2 / log (1.0064) = 108.650 years
