# Researchers use formula f (t) = P e^{r t} to model the population of fruit flies growing...

## Question:

Researchers use formula {eq}f (t) = P e^{r t} {/eq} to model the population of fruit flies growing exponentially after t days. Measurements show that {eq}f (0) = 30 {/eq} and {eq}f (5) = 1620 {/eq}.

a) Find a function of the form {eq}f(t) = P e^{rt} {/eq} that models the fruit fly population after {eq}t {/eq} days.

b) When the function f is expressed in the {eq}f (x) = Pa^x {/eq}, the constant {eq}a {/eq} is the growth factor. Find the growth factor.

c) When will there be {eq}18,000 {/eq} fruit flies?

## Exponential Growth

Exponential growth describes the increase of a quantity over time, wherein the derivative of the quantity is directly proportional to the quantity itself. It is commonly used to model the growth behavior of microorganisms and insects.

## Answer and Explanation:

a) To determine the corresponding function, we must calculate the value of P and r using the given boundary conditions.

{eq}f (0) = 30\\ f (5) = 1620\\ {/eq}

{eq}f(0) =P e^{r(0)}=30\\ Pe^0=30\\ P=30\\ {/eq}

{eq}f(5)=30e^{r(5)}=1620\\ 30e^{5r}=1620\\ e^{5r}=\dfrac{1620}{30}\\ \ln \mid e^{5r} \mid=\ln \mid 54 \mid\\ 5r\approx 3.99\\ r\approx \dfrac{3.99}{5}\\ r\approx 0.8 {/eq}

Therefore, the exponential equation would be given by

{eq}\color{red}{f (t) = 30 e^{0.8t}}\\ {/eq}

b) Using the same set of boundary conditions, we can the determine the exponential equation in the following form

{eq}f (x) = Pa^x {/eq}

{eq}f(0)=Pa^(0)=30\\ P=30\\ {/eq}

{eq}f(5)=30a^5=1620\\ a^5=\dfrac{1620}{30}\\ a= (54)^{1/5}\\ \color{blue}{a\approx 2.22}\\ {/eq}

The equation becomes

{eq}\color{red}{f(x)=(30)(2.22)^{x}}\\ {/eq}

c) We solve for the corresponding time for P = 18,000 using both equations.

{eq}30 e^{0.8t}=18000\\ e^{0.8t}=\dfrac{18000}{30}\\ \ln \mid e^{0.8t} \mid =\ln \mid 600 \mid\\ 0.8t= 6.40 \\ t= \dfrac{6.4}{0.8}\\ \color{blue}{t = 8\,\rm days}\\[0.2cm] {/eq}

{eq}(30)(2.22)^{x}=18000\\ 2.22^x=\dfrac{18000}{30}\\ \ln \mid2.22^x \mid= \ln \mid 600 \mid\\ x\ln \mid2.22 \mid= 6.4\\ x= \dfrac{6.4}{\ln \mid2.22 \mid}\\ \color{blue}{x= 8\,\rm days} {/eq}

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10