Question

The doubling period of a bacterial population is 15 minutes. At time t=110 minutes, the bacterial population was 50000.

What was the initial population at time t=0?

Find the size of the bacterial population after 5 hours.

Answer 1

The initial population is about 310 bacteria.

The size of the bacterial population after five hours is about 325, 058, 560.

Explanation:

The exponential function describing growth is given by:

`\displaystyle A=A_0(r)^(t/d)`

Where A is the amount afterwards, A₀ is the initial amount, r is the rate of growth, t is the amount of time that has passed and d is the time it takes for one cycle (both t and d are minutes in this case).

We are given that the bacteria population doubles after every 15 minutes. Hence, r = 2 and d = 15:

`\displaystyle A=A_0\left(2\right)^(t/15)`

At time t = 110 minutes, the bacterial population was 50,000. Hence:

`50000=A_0(2)^(110/15)`

Solve for A₀:

`\displaystyle 50000=A_0(2)^(22/3)\Rightarrow \displaystyle A_0=(50000)/(2^(22/3))=310.0392...\approx310`

Hence, the initial population (when t = 0) was about 310.

Thus, our function is:

`A=310 (2)^(t/15)`

After five hours, 300 minutes would have passed. Thus:

`A(300)=310(2)^(300/15)=310(2)^(20)=325, 058, 560\text{ bacteria}`