# 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.

## Answer and Explanation:

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

{eq}P(t)=Ae^{kt} \qquad (1) {/eq}

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.**

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10