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.**

**Expert Answer**

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 $2018$.