Question

If 3,000 bacteria, with a growth constant (k) of 2.8 per hour, are present at the beginning of an experiment, in how many hours will there be 15,000 bacteria?

Answer 1

Given:

Initial number of bacteria = 3000

With a growth constant (k) of 2.8 per hour.

To find:

The number of hours it will take to be 15,000 bacteria.

Solution:

Let P(t) be the number of bacteria after t number of hours.

The exponential growth model (continuously) is:

`P(t)=P_0e^(kt)`

Where,
`P_0` is the initial value, k is the growth constant and t is the number of years.

Putting
`P(t)=15000,P_0=3000, k=2.8` in the above formula, we get

`15000=3000e^(2.8t)`

`(15000)/(3000)=e^(2.8t)`

`5=e^(2.8t)`

Taking ln on both sides, we get

`\ln 5=\ln e^(2.8t)`

`1.609438=2.8t`
`[\because \ln e^x=x]`

`(1.609438)/(2.8)=t`

`0.574799=t`

`t\approx 0.575`

Therefore, the number of bacteria will be 15,000 after 0.575 hours.