The aim of this question is to understand the **exponential function,** how to fit the **points** into the **exponent model** and understand what the exponential function describes.

In mathematics, the exponential function is described by a relation of the **form** **y=a^x**. where the **independent** variable **x** goes over the entire **real number** and **a** is a constant number that is greater than zero. **a** in **exponential function** is known as the base of the function. **y=e^x** or **y=exp(x)** is one of the most important **exponential function** where the **e** is **2.7182818**, base of the natural system of **logarithms** **(ln)**

An exponential model **grows** or **decays** depending on the function. In exponential **growth** or exponential **decay**, an amount **rises** or **falls** by an appointed percent at regular intervals.

In exponential growth, the **quantity** rises slowly but **increases** rapidly after some intervals. As time passes, the rate of change becomes **faster.** This change in **growth** is marked as an **exponential increase.** The **formula** for exponential growth is denoted by:

\[y = a(1+r)^x \]

where $r$ **represents** the growth rate.

In exponential decay, The quantity **falls** rapidly at first but **slows** down after some **intervals.** As time passes, the rate of change becomes **slower.** This change in growth is marked as an **exponential decrease.** The **formula** for exponential decay is denoted by:

\[y = a(1-r)^x \]

where $r$ **represents** the decay percentage.

## Expert Answer

Given **points** are $(0,8)$ and $(1,3)$.

General **equation** of the exponential **model** is $y = ae^{bx}$.

So first we will take the point $(0,8)$ and **substitute** in the general equation and **solve** for $a$.

**Inserting** the $(0,8)$ in the general equation will **eliminate** $b$ as it will get **multiplied** by $0$ and hence will make it easy to **solve** for $a$:

\[y = ae^{bx}\]

Inserting $(0,8)$:

\[8 =ae^{b(0)}\]

\[8 =ae^0\]

Anything with **power** $0$ is $1$, so:

\[a =8\]

Now that the $a$ is known, **Insert** the point $(1,3)$ and solve for $b$:

\[y=ae^{bx}\]

\[3=ae^{b(1)}\]

Inserting $a=8$:

\[3=8e^{b}\]

\[e^b=\dfrac{3}{8}\]

Taking $ln$ to solve for $b$:

\[b= ln(\dfrac{3}{8})\]

## Numerical Answer

**Exponential model** that fits the points $(0,8)$ and $(1,3)$ is $y = 8e^{ln \left(\dfrac{3}{8}\right) } $.

## Example

How do you find the **exponential model** $y=ae^{bx}$ that fits the two **points** $(0, 2)$, $(4, 3)$?

Given **points** are $(0,2)$ and $(4,3)$.

**Exponential** model in the **question** is given as $y = ae^{bx}$.

So first we will **plug** in the point $(0,8)$ in the **general equation** and solve for $a$.

Reason for **plugging** this point that by **inserting** $(0,8)$ in the given **equation,** it will **eliminate** $b$ and hence will make it easy to **solve** for $a$.

\[y=ae^{bx}\]

Inserting $(0,2)$:

\[2=ae^{b(0)}\]

\[2=ae^0\]

Anything with **power** $0$ is $1$ so:

\[a =2\]

Now that the $a$ is **known,** Insert the point $(4,3)$ and **solve** for $b$.

\[ y=ae^{bx} \]

\[3=ae^{b(4)}\]

Inserting $a=2$:

\[3= 2e^{4b}\]

\[e^{4b}= \dfrac{3}{2}\]

Taking $ln$ to solve for $b$:

\[ 4b= ln(\dfrac{3}{2}) \]

\[ b= \dfrac{ln(\dfrac{3}{2})}{4} \]

**Exponential** model that fits the **points $y=2e^{101x}$** $(0,2)$ and $(4,3)$ is **$y = 2e^{0.101x}$**.