Answer
a)
The required expression is ${{P}_{t}}={{P}_{0}}{{e}^{0.031t}}$.
b)
The balance after 1 year is $\$1031.49$ approximately and the balance after 2 years is $\$1063.96$ approximately.
c)
The doubling time is about $\text{22}\text{.4 years}$.
Work Step by Step
a)
The exponential growth model $P\left( t \right)={{P}_{0}}{{e}^{kt}},k>0$.
Here, ${{P}_{0}}$ is the investment at time 0, $P\left( t \right)$ is the balance at time $t$ years, $k$ is the interest rate compounded continuously, and doubling time is the amount of time necessary for the investment to double in size.
It is given that the interest rate is $3.1%$ per year compounded continuously.
Thus, $k=0.031$.
Substitute $k=0.031$ in $P\left( t \right)={{P}_{0}}{{e}^{kt}}$:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{kt}} \\
& ={{P}_{0}}{{e}^{0.031t}}
\end{align}$
b)
Note that the expression for the balance after $t$ years is $P\left( t \right)={{P}_{0}}{{e}^{0.031t}}$ from part a).
Thus, substitute ${{P}_{0}}=1000$ and $t=1$ in $P\left( t \right)={{P}_{0}}{{e}^{0.031t}}$ to calculate the balance after 1 year as follows:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{0.031t}} \\
& P\left( 1 \right)=1000{{e}^{0.031\left( 1 \right)}} \\
& =1000{{e}^{0.031}}
\end{align}$
Using a calculator, ${{e}^{0.031}}\approx 1.031485$. So,
$\begin{align}
& P\left( 1 \right)=1000{{e}^{0.031}} \\
& \approx 1000\times 1.031485 \\
& \approx 1031.485 \\
& \approx 1031.49
\end{align}$
Therefore, the balance after 1 year is $\$1031.49$ approximately.
Similarly, to calculate the balance after 2 years, substitute ${{P}_{0}}=1000$ and $t=2$ in $P\left( t \right)={{P}_{0}}{{e}^{0.031t}}$:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{0.031t}} \\
& P\left( 2 \right)=1000{{e}^{0.031\left( 2 \right)}} \\
& =1000{{e}^{0.062}}
\end{align}$
$\begin{align}
& P\left( 2 \right)=1000{{e}^{0.031}} \\
& \approx 1000\times 1.063962 \\
& \approx 1063.962 \\
& \approx 1063.96
\end{align}$
c)
Doubling time is the amount of time necessary for the investment to double in size.
If ${{P}_{0}}=\$1000$ is the investment at time 0, then the double of ${{P}_{0}}$ is :
$\begin{align}
& 2{{P}_{0}}=2\times \$1000\\&=\$2000\end{align}$.
Thus, to find the doubling time, replace $P\left( t \right)$ with $2000$ and ${{P}_{0}}=\$1000$ in $P\left( t \right)={{P}_{0}}{{e}^{0.031t}}$; substitute and solve for $t$:
$\begin{align}
& P\left( t \right)=1000{{e}^{0.031t}} \\
& 2000=1000{{e}^{0.031t}} \\
\end{align}$
Divide by $5000$ on both the sides:
$\begin{align}
& \frac{2000}{1000}=\frac{1000{{e}^{0.031t}}}{1000} \\
& 2={{e}^{0.031t}} \\
\end{align}$
Take natural logarithm on both the sides:
$\ln 2=\ln {{e}^{0.031t}}$
Apply the power rule of logarithms $\ln {{m}^{n}}=n\ln m$ in $\ln {{e}^{0.031t}}$:
$\begin{align}
& \ln 2=\ln {{e}^{0.031t}} \\
& \ln 2=0.031t\operatorname{lne} \\
\end{align}$
$\begin{align}
& \ln 2=0.031t\ln e \\
& 0.6931=0.031t
\end{align}$
Divide by $0.031$ on both sides,
$\begin{align}
& \frac{0.6931}{0.031}=\frac{0.031t}{0.031} \\
& 22.35\approx t \\
& 22.4\approx t
\end{align}$
