# Frogs have been breeding like flies at the Enormous State University (ESU) campus! Every year,...

## Question:

Frogs have been breeding like flies at the Enormous State University (ESU) campus! Every year, the pledge class of the Epsilon Delta fraternity is instructed to tag all the frogs residing on the ESU campus. Two years ago {eq}(t = 0) {/eq} they managed to tag all {eq}46,000 {/eq} of them (with little Epsilon Delta Fraternity tags). This year's pledge class discovered that all the tags had all fallen off, and they wound up tagging a total of {eq}55,200 {/eq} frogs.

a) Find an exponential model for the frog population.

b) Assuming exponential population growth, and that all this year's tags have fallen off, how many tags should Epsilon Delta order for next year's pledge class?

## Exponential Growth:

Suppose that a quantity {eq}Q(t) {/eq} obeys an equation of the form {eq}Q(t)=AB^t {/eq} for some constants {eq}A>0 {/eq} and {eq}B>1 {/eq}. Then we say that the quantity {eq}Q(t) {/eq} grows exponentially.

Exponential growth generally occurs when the quantity is growing without restriction. For example, a population which has no constraints (has enough food, enough space, and so on) may grow exponentially.

a) Let {eq}P(t) {/eq} be the population of frogs after {eq}t {/eq} years, where {eq}t=0 {/eq} corresponds to the time {eq}2 {/eq} years ago. Then we're given that {eq}P(0)=46000 {/eq} and {eq}P(1)=55200 {/eq}.

If the population is growing exponentially, then {eq}P(t)=AB^t {/eq} for some constants {eq}A {/eq} and {eq}B {/eq}. Substituting the above values into this formula gives:

{eq}\begin{align*} 46000&=P(0)\\ &=AB^0\\ &=A \end{align*} {/eq}

and

{eq}\begin{align*} 55200&=P(1)\\ &=AB^1\\ &=AB\\ &=46000B&&\text{(since }A=46000\text{ as above)}\\ \frac{55200}{46000}&=B\\ \frac{6}{5}&=B \, . \end{align*} {/eq}

So an exponential model for the population growth is {eq}\boxed{P(t)=46000\left(\frac{6}{5}\right)^t}\, {/eq}.

b) If all the tags have fallen off, the pledges need to tag the entire population of frogs, and the current year is year {eq}t=2 {/eq}. Using the above exponential model, this means that the number of frogs they will need to tag is:

{eq}\begin{align*} P(2)&=46000\left(\frac{6}{5}\right)^2\\ &=66240 \, . \end{align*} {/eq}

So the fraternity should order {eq}\boxed{66240} {/eq} tags (or maybe a few more, just to be on the safe side).