Question

A culture of 100 bacteria grows at a continuous rate of 20%. Two hours later. 60 of the same type of bacteria are placed in a culture that allows a 30% growth rate. After how many hours do both cultures have the same population? Explain your answer.

Answer 1

Both cultures will have the same population after 11 hours.

Explanation:

Exponential equation for population growth:

The exponential equation for population growth is given by:

`P(t) = P(0)(1+r)^t`

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

A culture of 100 bacteria grows at a continuous rate of 20%.

This means that
`P(0) = 100, r = 0.2`. So

`P(t) = P(0)(1+r)^t`

`P_(1)(t) = 100(1+0.2)^t`

`P_(1)(t) = 100(1.2)^t`

The other culture starts to hours later, and we want to take that time as a parameter. So

`P_(1)(2) = 100(1.2)^2 = 144`

Which means that when culture 2 growth starts, the population size of culture 1 is described by the following equation:

`P_(1)(t) = 144(1.2)^t`

Two hours later 60 of the same type of bacteria are placed in a culture that allows a 30% growth rate.

This means that
`P(0) = 60, r = 0.3`. So

`P(t) = P(0)(1+r)^t`

`P_(2)(t) = 60(1+0.3)^t`

`P_(2)(t) = 60(1.3)^t`

After how many hours do both cultures have the same population?

This is t for which:

`P_(1)(t) = P_(2)(t)`

So

`144(1.2)^t = 60(1.3)^t`

`((1.3)^t)/((1.2)^t) = (144)/(60)`

`((1.3)/(1.2))^t = (144)/(60)`

`\log{((1.3)/(1.2))^t} = \log{(144)/(60)}`

`t\log{(1.3)/(1.2)} = \log{(144)/(60)}`

`t = \frac{\log{(144)/(60)}}{\log{(1.3)/(1.2)}}`

`t = 11`

Both cultures will have the same population after 11 hours.