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The Natural Base e

The Natural Base e 1.15 - Solution

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

A=Pert\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. A: amount in an accountP: principale: Euler’s numberr: annual interest rate written as a decimalt: number of years\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. M=3224e0.05t\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.\$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 10{\color{#009600}{10}} years by substituting t=10{\color{#009600}{t}}={\color{#009600}{10}} in the given equation.
M=3224e0.05tM=3224e^{0.05t}
M=3224e0.05(10)M=3224e^{0.05({\color{#009600}{10}})}
Simplify right-hand side
M=3224e0.5M=3224e^{0.5}
M=3224(1.64872)M=3224(1.64872\ldots)
M=5315.47737M=5315.47737\ldots
M5315.48M\approx 5315.48
Peter's savings after 1010 years is about $5315.48.\textcolor{darkorange}{\$5315.48}. The graph also shows that the balance is about $7250\textcolor{darkorange}{\$7250} when t=10.t=10. Thus, Patricia has more money after 1010 years.