Differential Equations: Population of a city in 1990

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agentcollins
Differential Equations: Population of a city in 1990

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
Jhun Vert's picture

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
Jhun Vert's picture

$x = x_o e^{kt}$

$1,000,000 = \frac{1,000,000}{3}e^{k(2010 - 2006)}$

$3 = e^{4k}$

$e^k = 3^{1/4}$
 

$x = x_o(3^{t/4})$

$1,000,000 = \Big[ 3^{(2010 - 1990)/4} \Big] x_o $

$1,000,000 = 3^5x_o$

$x_o = 4115.23$       answer

Jhun Vert
Jhun Vert's picture

Casio fx-991ES PLUS
Mode3:Stat5: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 = Shift1:Stat5:Reg5:ŷ

agentcollins

Wow. This is so helpful. Don't you make subject topics regarding shortcut tricks in scientific calculator? Can I use all regression equations for elementary de application problem solving?

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