# The doubling period of a bacterial population is 20 minutes. At time t = 100 minutes, the...

## Question:

The doubling period of a bacterial population is 20 minutes. At time {eq}t= {/eq} 100 minutes, the bacterial population was 80,000. What was the initial population at time {eq}t=0 {/eq} ? Find the size of the bacterial population after 5 hours.

## Model of Exponential Growth:

A bacterial population doubles in a given amount of time. Then we are also given the size of the bacterial population at a different point in time. Using all the information given in the question we calculate the initial bacterial population and then the bacterial population at a given point of time in the future. The concepts used in this question include solving an exponential equation using natural logarithms.

Using an exponential function to model the population of bacterial we have the population P at time t in minutes given by

where A = initial population and k = constant to be determined.

Then using the fact that the bacterial population doubles in 20 minutes we let t = 20 in (1) to obtain

{eq}2A=Ae^{k(20)} \implies 2 = e^{20k} \qquad (2) {/eq}

Taking natural logarithms of both sides of (2) yields

{eq}\displaystyle \ln 2 = 20 k \implies k = \frac {\ln 2}{20} \qquad (3) {/eq}

Substituting the value of k from (3) in (1) gives

{eq}\displaystyle P(t)=Ae^{\frac {\ln2}{20}t} \implies P(t) = A \left( 2^{t/20} \right) \qquad (4) {/eq}

The question also gives us that the P(t) = 80000 when t = 100 minutes. Using this information in (4) yields

{eq}\displaystyle 80000=A \left( 2^{100/20} \right) \implies 80000 = A \left( 2^5 \right) \implies A = \frac {80000}{32} =2500. {/eq}

Since A is the initial population, hence the initial population at time t = 0 minutes is 2500 bacteria.

Substituting A = 2500 in (4) gives us our complete bacterial population model which will be

{eq}\displaystyle P(t) = 2500 \left( 2^{t/20} \right) \qquad (5) {/eq}

So after 5 hours = 300 minutes the bacterial population from (5) will be

{eq}\displaystyle P(300) = 2500 \left( 2^{300/20} \right) = 2500 \left( 2^{15} \right) = 81920000. {/eq}

Therefore the bacterial population after 5 hours will be 81920000 bacteria.