# Suppose an E. coli culture is growing exponentially at { 37^o/C }. After 20 minutes at that...

## Question:

Suppose an E. coli culture is growing exponentially at {eq}37^o/C {/eq}. After 20 minutes at that temperature, there are {eq}1.28 \times10^7 {/eq} E. coli cells. After 60 minutes, there are {eq}2.4 \times 10^7 {/eq} cells.

How long does it take for the culture to have double the amount of cells that it had started with?

## Exponential Growth :

Organisms such as bacteria which grow at a rate proportional to their population size, are generally modeled using an exponential model. If we know the population at two different times, we can find the rate of growth of the population and thus complete our model.

## Answer and Explanation:

As the E.coli culture is growing exponentially, we can use an exponential model of the kind given below to model its population.

{eq}P(t) =P(o) e^{kt} {/eq}

Here, **P(o)** is the initial population and **P(t)** is the population at time **t** with **k** being the rate of growth.With the information provided about the growth of the culture, we can fid the rate of growth per minute.

We know the population after 20 mintues and we know the population after 40 minutes after 20 minutes.

{eq}\Rightarrow P(40) =1.28*10^7e^{40k}=2.4*10^7\\ e^{40k}=\frac {2.4}{1.28}\\ 40k=\ln 1.875\\ k\approx 0.015715 \\\therefore P(t) =P(o) e^{0.015715t} {/eq}

Having found the rate of growth, we can find the population's doubling time as follows.

{eq}2P(o)=P(o)e^{0.015715t} \\ \ln 2=0.015715t\\ t\approx 44\,\text {minutes} {/eq}

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10