# There were 60 flies after the 2nd day of the experiment and 360 flies after the fourth day. How...

## Question:

There were 60 flies after the 2nd day of the experiment and 360 flies after the fourth day. How many flies were there in the original population?

## Exponential Growth Model: Population

A population growth is commonly described by an exponential growth function.

Given:

- An initial population {eq}p_o {/eq}

- A population at time {eq}t {/eq}, {eq}p_t {/eq}

- Relative growth rate {eq}k {/eq}

We can write a population function as {eq}p_t = p_oe^{kt} {/eq}

## Answer and Explanation:

Suppose that our exponential function of this population is:

{eq}p_t = p_oe^{kt} {/eq}

We have to identify two constants:

{eq}p_o {/eq} and {eq}k {/eq}

Given two points:

{eq}(t_1, p_1) = (2, 60)\\ (t_2, p_2) = (4, 360) {/eq}

We get two equations:

{eq}60 = p_oe^{2k}~~~(1)\\ 360 = p_oe^{4k}~~~(2) {/eq}

Let's divide equation (2) by equation (1):

{eq}6 = e^{4k - 2k} = e^{2k} {/eq}

Solving for the growth rate:

{eq}\ln(6) = 2k\\ k = \dfrac{1}{2}\ln(6) {/eq}

Now, let's use equation (1) to solve for the initial population:

{eq}p_o = p_1e^{-kt_1}\\ p_o = 60e^{-\dfrac{1}{2}\ln(6)\cdot 2} = 10 {/eq}

Our initial population, as a result, is {eq}\boxed{p_o = 10~\rm{flies}} {/eq}

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10