# The fox population in a certain region has an annual growth rate of 9 percent per year. It is estimated that the population in the year 2010 was 23,900. Find a function for the population and estimate the fox population in the year 2018.

This article aims to find the population growth. Exponential growth is the process that increases quantity over time. It occurs when instantaneous rate of change (i.e., derivative) of an amount with respect to time is proportional to the quantity itself. A quantity undergoing exponential growth is an exponential function of time; that is, the variable representing time is an exponent (unlike other types of growth, such as quadratic growth).

If the proportionality constant is negative, then the quantity decreases over time and is said to undergo exponential decay. A discrete region of definition with equal intervals is also called geometric growth or geometric decrease since the function values form geometric progression.

Exponential growth is a data pattern that shows an increase over time by creating an exponential function curve. For example, suppose the cockroach population grows each year exponentially, starting with $3$ in first year, then $9$ in second year, $729$ in third year, and $387420489$ in the fourth year, and so on. The population, in this case, grows every year to the power of $3$. The exponential growth formula, as its name suggests, involves exponents. Exponential growth models include several formulas.

Formula $1$

$f(x)=x_{o}(1+r)^{t}$

Formula $2$

$f(x)=ab^{x}$

Formula $3$

$A=A_{o}e^{kt}$

Where $A_{o}$ is the initial value.

$r$ is the rate of growth.

$k$ is the constant of proportionality.

The growth of a bacterial colony is often used as an illustration. One bacterium divides into two, each of which divides, resulting in four, then eight, $16$, $32$, and so on. The amount of growth keeps increasing because it is proportional to the ever-increasing number of bacteria. Growth like this is seen in real-life activities or phenomena, such as the spread of a viral infection, the growth of debt due to compound interest, and the spread of viral videos.

Given that it is an exponential growth problem.

The exponential growth is expressed as,

$A_{t}=A_{o}e^{kt}$

$A_{t}$ is the population at $t$.

$A_{o}$ is the initial population.

$k$ is the growth constant.

$t$ is the time.

Let $X$ be the initial population growing at $9\%$, given the initial time in $2010$ and the final time in $2018$; our population is estimated to be:

$A_{t}=23900e^{2018-2010}K$

$=23900e^{8\times 0.09}$

$=49101$

$A_{t}=49101$

Hence, the fox population is estimated as $49,101$ in $2018$.

## Numerical Result

The fox population is estimated to be $49,101$ in $2018$.

## Example

The fox population in a particular area has an annual growth rate of $10\:percent$ per year. It had an estimated population of $25000$ in $2010$. Find the population function and estimate the fox population in $2018$.

Solution

Given that it is an exponential growth problem.

The exponential growth is expressed as,

$A_{t}=A_{o}e^{kt}$

$A_{t}$ is the population at $t$.

$A_{o}$ is the initial population.

$k$ is the growth constant.

$t$ is the time.

Let $X$ be the initial population growing at $10\%$, given the initial time in $2010$ and the final time in $2018$; our population is estimated to be:

$A_{t}=25000e^{2018-2010}K$

$=25000e^{8\times 0.1}$

$=55,638$

$A_{t}=55,638$

Hence, the fox population is estimated as $55,638$ in $2018$.