# Suppose that the population of deer in a state is 19,900 and is growing 3 percentage each year....

## Question:

Suppose that the population of deer in a state is {eq}19,900 {/eq} and is growing {eq}3 \% {/eq} each year.

Predict the population after {eq}10 {/eq} years.

## Exponential Growth Function:

Knowing that the equation of an exponential growth function is

$$f(x) = a \cdot (1+ r)^t$$

where {eq}a {/eq} is the initial population, {eq}r {/eq} is the rate, and {eq}t {/eq} is the term position, one can find the population in ten years by plugging in the current population, percentage growth, and years into the equation.

Given:

• {eq}a = 19,900 {/eq}
• {eq}r = 3\% = 0.03 {/eq}
• {eq}t = 10 {/eq}

Using the formula for exponential growth, the population of deer after {eq}10{/eq} years is calculated as follows:

\begin{align} f(x) &= (19,900)\cdot (1 + (0.03))^{(10)}\\[0.3cm] &= 19,900 \cdot (1.03)^{10}\\[0.3cm] &= 19,900 \cdot 1.343916379\\[0.3cm] &= 26,743.93595 \\[0.3cm] &\approx 26,744 \text{ deer} \end{align} \\

In ten years, the deer population will be approximately {eq}26,744{/eq} deer-strong. 