# Differential Equations: Population of a city in 1990

Submitted by agentcollins on July 26, 2016 - 10:33pm

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.

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## 2010 = 1,000,000

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## $x = x_o e^{kt}$

$x = x_o e^{kt}$

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

$3 = e^{4k}$

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

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

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

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

$x_o = 4115.23$

answer## Casio fx-991-ES PLUSMode → 3

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

Note: 2010 = 20 years after 1990

AC

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

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

## Wow. This is so helpful. Don

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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