Question

4 years ago, the population of a city was of "x" inhabitant, 2 years later, that is to say two years ago, the population of this same city was 81,000 inhabitants and today it is 65,610. Using this data, find the population of four years ago.

Answer 1

Answer:

The population of four years ago was 100,783 inhabitants

Step-by-step explanation:

The population of the city after t years is given by the following equation:

$P(t) = P(0)(1-r)^{t}$

In which P(0) is the initial population and r is the decrease rate, as a decimal.

2 years later, that is to say two years ago, the population of this same city was 81,000 inhabitants and today it is 65,610.

This means that:

$P(2) = 81000, P(4) = 65610$

We are going to use this to build a system, and find P(0), which is the initial population(four years ago).

P(2) = 81000

$P(t) = P(0)(1-r)^{t}$

$81000 = P(0)(1-r)^{2}$

$(1-r)^{2} = \frac{81000}{P(0)}$

P(4) = 65610

$P(t) = P(0)(1-r)^{t}$

$65100 = P(0)(1-r)^{4}$

$65100 = P(0)((1-r)^{2})^{2}$

Since $(1-r)^{2} = \frac{81000}{P(0)}$

$65100 = P(0)(\frac{81000}{P(0)})^{2}$

Using P(0) = x

$65100 = x(\frac{81000}{x})^{2}$

$65100 = \frac{6561000000x}{x^{2}}$

$65100x^{2} = 6561000000x$

$65100x^{2} - 6561000000x$

$x(65100x - 6561000000) = 0$

x = 0, which does not interest us, or:

$65100x - 6561000000 = 0$

$65100x = 6561000000$

$x = \frac{6561000000}{65100}$

$x = 100,783$

The population of four years ago was 100,783 inhabitants