# ans this questionThe population of a certain community is known to increase at a rate proportional to the number of people present at any time. The population grows from 100 people to 500 people...

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The population of a certain community is known to increase at a rate proportional to the number of people present at any time. The population grows from 100 people to 500 people in 5 years. How long will it take for the original population to double? What will be the population in 30 years?

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You know that the rate of growth of population is said to be proportional to the number of people at any time.

Put this statement into an equation such that dP/dt = nP

P denotes the population and t denotes the time.

n denotes a constant.

Divide the equation above by P:

`(dP)/(P*dt) = n`

Multiply this equation by dt:

`(dP)/P = n*dt`

`` Integrating both sides yields: `int (dP)/P = int n*dt`

``

`ln P = n*t + c =gt` The function `P(t) = c*e^(n*t)`

If `t = 0, P = 100 =gt P(0) = 100 lt=gt 100 = c*e^0 = c`

If the constant c = 100, the equation becomes `P(t) = 100e^(n*t)`

Solve for n the equation above if t = 5 and P = 500.

`500 = 100e^(5n)`

Dividing by 100 yields: `5 = e^(5n)`

Taking logarithm yields `ln 5 = 5n =gt n = (ln5)/5`

This equation that shows how the population grows is:

`P(t) = 100e^((t*ln5)/5)`

If the original population doubles then P = 200.

`200 = 100e^((t*ln5)/5) =gt 2 = e^((t*ln5)/5) =gt ln 2 = t*(ln 5)/5`

`=gt t = (5*ln2)/(ln 5) =gt t = (ln 2^5)/(ln 5) = (ln 32)/(ln 5)`

`t = 3.465/1.609 ~~ 2 years`

It will take about 2 years for the population to double.

Let's evaluate the population in 30 years.

`P(30) = 100e^((30*ln5)/5)`

`P(30) = 100*e^(ln 5^6) =gt P(30) = 100*5^6 = 1562500`

**It will take about 2 years for the population to double and the population will reach the number of 1562500 in 30 years.**