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# The Natural Base e

## The Natural Base e 1.15 - Solution

a
Let's begin with Peter's saving. We will first examine the Continuously Compounded Interest model.

$\begin{gathered} \textcolor{darkorange}{A}={\color{#0000FF}{P}}e^{\textcolor{magenta}{r}{\color{#009600}{t}}} \end{gathered}$ Its components can be defined as shown below. \begin{aligned} \textcolor{darkorange}{A}\text{:}& \text{ amount in an account}\\ {\color{#0000FF}{P}}\text{:}& \text{ principal}\\ e\text{:}& \text{ Euler's number}\\ \textcolor{magenta}{r}\text{:}& \text{ annual interest rate written as a decimal}\\ {\color{#009600}{t}}\text{:}& \text{ number of years} \end{aligned} With this, we can analyze the equation. $\begin{gathered} \textcolor{darkorange}{M}={\color{#0000FF}{3224}}e^{\textcolor{magenta}{0.05}{\color{#009600}{t}}} \end{gathered}$ Therefore, the principal of the Peter's savings is $\3224.$ Next, let's examine the graph that models Patricia's savings.

Therefore, Patricia started with more money.

b
We can also find the balance of the savings after ${\color{#009600}{10}}$ years by substituting ${\color{#009600}{t}}={\color{#009600}{10}}$ in the given equation.
$M=3224e^{0.05t}$
$M=3224e^{0.05({\color{#009600}{10}})}$
Simplify right-hand side
$M=3224e^{0.5}$
$M=3224(1.64872\ldots)$
$M=5315.47737\ldots$
$M\approx 5315.48$
Peter's savings after $10$ years is about $\textcolor{darkorange}{\5315.48}.$ The graph also shows that the balance is about $\textcolor{darkorange}{\7250}$ when $t=10.$ Thus, Patricia has more money after $10$ years.