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# The Natural Base e

## The Natural Base e 1.16 - Solution

a
If an element decays continuously, we can use the following equation to model the situation.

$\begin{gathered} \textcolor{darkorange}{y}={\color{#0000FF}{I}}e^{\textcolor{magenta}{r}{\color{#009600}{t}}} \end{gathered}$ Its components can be defined as shown below. \begin{aligned} \textcolor{darkorange}{y}\text{:}& \text{ remaining amount}\\ {\color{#0000FF}{I}}\text{:}& \text{ initial amount}\\ e\text{:}& \text{ Euler's number}\\ \textcolor{magenta}{r}\text{:}& \text{ decay rate}\\ {\color{#009600}{t}}\text{:}& \text{ number of years} \end{aligned} With this, we can analyze the equation that models the tritium decay. $\begin{gathered} \textcolor{darkorange}{y}={\color{#0000FF}{10}}e^{\textcolor{magenta}{\text{-}0.0562}{\color{#009600}{t}}} \end{gathered}$ Therefore, the initial amount of uranium is $10$ milligrams. Looking at the graph, we can read the initial value as $15.$

Radium has the highest amount in the beginning.

b
We can also find the remaining amount after ${\color{#009600}{10}}$ years by substituting ${\color{#009600}{t}}={\color{#009600}{10}}$ in the given equation.
$y=10e^{\text{-}0.0562t}$
$y=10e^{\text{-}0.0562({\color{#009600}{10}})}$
Simplify right-hand side
$y=10e^{\text{-}0.562}$
$y=10(0.57006\ldots)$
$y=5.7006\ldots$
$y\approx5.70$
The initial amount of uranium is about $\textcolor{darkorange}{5.70}$ milligrams. Next, let's examine the graph that models the radium decay.

The graph also shows that the remaining amount is about $\textcolor{darkorange}{1}$ milligram when $t=10.$ Therefore, uranium has a greater remaining amount.