Question

An experiment starts and the population of a bacteria culture increases by 10% in the first hour. Assume that the population increases at a rate proportional to the amount of bacteria.

a) What is the doubling time?
b) If the initial population is 2,000, what is the population in 6 hours?

Answer 1

a) The doubling time is 7.27 hours.

b) The population is 6 hours will be 3,543.

Explanation:

The population of the bacteria is modeled by the following equation:

`P(t) = P(0)e^(rt)`

In which P(0) is the initial population, P(t) is the population after t hours and r is the growth rate.

An experiment starts and the population of a bacteria culture increases by 10% in the first hour.

This means that

`P(1) = 1.1P(0)`

Which helps us find r.

So

`P(t) = P(0)e^(rt)`

`1.1P(0) = P(0)e^(r)`

`e^(r) = 1.1`

Applying ln to both sides

`\ln{e^(r)} = ln(1.1)`

`r = 0.0953`

So

`P(t) = P(0)e^(0.0953t)`

a) What is the doubling time?

This is t when
`P(t) = 2P(0)`

So

`P(t) = P(0)e^(0.0953t)`

`2P(0) = P(0)e^(0.0953t)`

`e^(0.0953t) = 2`

Applying ln to both sides

`\ln{e^(0.0953t)} = ln(2)`

`0.0953t = ln(2)`

`t = (ln(2))/(0.0953)`

`t = 7.27`

The doubling time is 7.27 hours.

b) If the initial population is 2,000, what is the population in 6 hours?

This is P(6) when
`P(0) = 2000`. So

`P(t) = P(0)e^(0.0953t)`

`P(6) = 2000e^(0.0953*6) = 3543`

The population is 6 hours will be 3,543.