# In a research experiment, a population of fruit flies is increasing according to the law of...

## Question:

In a research experiment, a population of fruit flies is increasing according to the law of exponential growth. After 2 days there are 100 flies, and after 4 days there are 300 flies. How many flies will there be after 5 days?

## Exponential Functions

Exponential functions such as exponential growth and exponential decay functions can be used to describe physical and biological quantities such as population of organisms (exponential growth) and radioactive decay (exponential decay). The inverse of process of exponentiation is taking the logarithm.

## Answer and Explanation:

An exponential growth function given by {eq}\displaystyle y(t) = y_0 e^{kt} {/eq} is a function that models population growth where k is the growth constant, {eq}y_0 {/eq} is the initial population of the organisms and y(t) is the population of the organisms at any time, t.

For this problem, we have the following given:

{eq}y(2\ \text{days}) = 100\ \text{flies} {/eq}: number of flies after 2 days.

{eq}y(4\ \text{days}) = 300\ \text{flies} {/eq}: number of flies after 4 days.

To calculate the number of flies after 5 days, we need to write the exponential growth function for the flies by solving for {eq}y_0 {/eq} and k.

To solve for the constants, we can use substitution method. We write the two equations for the number of flies at {eq}t = 2\ \text{days} {/eq} and {eq}t = 4\ \text{days} {/eq}.

{eq}\displaystyle \begin{align*} 100\ \text{flies} &= y_0 e^{(2\ \text{days})k} \\ 300\ \text{flies} &= y_0 e^{(4\ \text{days})k} \\ \end{align*} {/eq}

We can rewrite the first equation and plug it in the second equation.

{eq}\displaystyle \begin{align*} 100\ \text{flies} &= y_0 e^{(2\ \text{days})k} \\ y_0 &= \frac{100\ \text{flies}} {e^{(2\ \text{days})k}}\\ y_0 &= (100\ \text{flies})e^{-(2\ \text{days})k} \end{align*} {/eq}

Plugging it in the second equation.

{eq}\displaystyle \begin{align*} 300\ \text{flies} &= y_0 e^{(4\ \text{days})k},\ y_0 = (100\ \text{flies})e^{-(2\ \text{days})k} \\ 300\ \text{flies} &= (100\ \text{flies})e^{-(2\ \text{days})k} e^{(4\ \text{days})k} \\ 300\ \text{flies} &=(100\ \text{flies})e^{(2\ \text{days})k}\\ \frac{300\ \text{flies}}{ (100\ \text{flies})}&=e^{(2\ \text{days})k}\\ e^{(2\ \text{days})k}&= 3 \end{align*} {/eq}

Taking the natural logarithm of both sides of the equation to remove the exponential and solve for k.

{eq}\displaystyle \begin{align*} \ln \bigg( e^{(2\ \text{days})k}&= 3\bigg) \\ (2\ \text{days})k&= \ln 3 \\ k&=\frac{1}{2} \ln 3 \text{ days}^{-1}\\ \end{align*} {/eq}

Now that we have k, we can plug it any of the two equations to solve for {eq}y_0 {/eq}.

{eq}\displaystyle \begin{align*} 100\ \text{flies} &= y_0 e^{(2\ \text{days})k} \\ 100\ \text{flies} &= y_0 e^{(2\ \text{days})\frac{1}{2} \ln 3 \text{ days}^{-1}} \\ y_0 &= \frac{100\ \text{flies} }{e^{(2\ \text{days})\frac{1}{2} \ln 3 \text{ days}^{-1}}}\\ &= \frac{100\ \text{flies} }{e^{\ln 3}}\\ y_0 &= \frac{100}{3} \end{align*} {/eq}

We now have an expression for the population of flies at any given day which is {eq}\displaystyle \boxed{ y(t) = \frac{100}{3} e^{(\frac{1}{2} \ln 3 \text{ days}^{-1}t)}} {/eq} where t is in days and y(t) is the number of flies.

To solve for the number of flies after 5 days, we plug {eq}t = 5\ \text{ days} {/eq} in {eq}y(t). {/eq}

Solving for {eq}y(5 \text{ days}) {/eq}:

{eq}\displaystyle \begin{align*} y(t) &= \frac{100}{3} e^{(\frac{1}{2} \ln 3 \text{ days}^{-1}t)}\\ y(5 \text{ days}) &= \frac{100}{3} e^{(\frac{1}{2} \ln 3 \text{ days}^{-1}(5\ \text{ days}))}\\ &\approx \boxed{ 520\ \text{flies}} \end{align*} {/eq}

Therefore, there will be {eq}\boxed{ 520\ \text{flies}} {/eq} after five days.

#### Learn more about this topic:

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10