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

$15000=3000e^{2.8t}$

$\dfrac{15000}{3000}=e^{2.8t}$

$5=e^{2.8t}$

Taking ln on both sides, we get

$\ln 5=\ln e^{2.8t}$

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

$\dfrac{1.609438}{2.8}=t$

$0.574799=t$

$t\approx 0.575$

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