# 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?

Jhun Vert 2010 = 1,000,000
2006 = 333,333.33
2002 = 111,111.11
1998 = 37,037.04
1994 = 12,345.68

Jhun Vert $x = x_o e^{kt}$

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

$3 = e^{4k}$

$e^k = 3^{1/4}$

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

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

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

$x_o = 4115.23$       answer

Jhun Vert 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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