Differential Equations: Population of a city in 1990

Submitted by agentcollins on

The topic is Population Growth, Decay and Investment. Somebody show the comple solution please.

A city has been found to have a population that triples every four years. If the city's population is one million in 2010, how many people were there in 1990?

The answer is 4,116.

Jhun Vert Thu, 07/28/2016 - 06:09

2010 = 1,000,000
2006 = 333,333.33
2002 = 111,111.11
1998 = 37,037.04
1994 = 12,345.68
1990 = 4,115.22       answer

Jhun Vert Thu, 07/28/2016 - 06:13

x=xoekt

1,000,000=1,000,0003ek(2010−2006)

3=e4k

ek=31/4
 

x=xo(3t/4)

1,000,000=[3(2010−1990)/4]xo

1,000,000=35xo

xo=4115.23       answer

Jhun Vert Thu, 07/28/2016 - 06:34

Casio fx-991ES PLUS
Mode → 3:Stat → 5:e^X

X Y
20 1,000,000
20-4 1,000,000 ÷ 3

Note: 2010 = 20 years after 1990

AC
Population in 1990 (zero year) = 0ŷ = 4115.23       answer

Note: ŷ = Shift → 1:Stat → 5:Reg → 5:ŷ