Answer
a)
The required expression is ${{P}_{t}}={{P}_{0}}{{e}^{0.025t}}$.
b)
The balance after 1 year is $\$5126.58$ approximately and the balance after 2 years is $\$5256.36$ approximately.
c)
The doubling time is about $\text{27}\text{.7 years}$.
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 $2.5\%$ per year compounded continuously.
Thus, $k=0.025$.
Substitute $k=0.025$ in $P\left( t \right)={{P}_{0}}{{e}^{kt}}$:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{kt}} \\
& P\left( t \right)={{P}_{0}}{{e}^{0.025t}}
\end{align}$
b)
Note that the expression for the balance after $t$ years is $P\left( t \right)={{P}_{0}}{{e}^{0.025t}}$ from part (a).
Thus, substitute ${{P}_{0}}=5000$ and $t=1$ in $P\left( t \right)={{P}_{0}}{{e}^{0.025t}}$ to calculate the balance after 1 year as follows:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{0.025t}} \\
& P\left( 1 \right)=5000{{e}^{0.025\left( 1 \right)}} \\
& =5000{{e}^{0.025}}
\end{align}$
Using a calculator, ${{e}^{0.025}}\approx 1.025315$. So,
$\begin{align}
& P\left( 1 \right)=5000{{e}^{0.025}} \\
& \approx 5000\times 1.025215 \\
& \approx 5126.575 \\
& \approx 5126.58
\end{align}$
Thus, the balance after 1 year is $\$5126.58$ approximately.
Similarly, to calculate the balance after 2 years, substitute ${{P}_{0}}=5000$ and $t=2$ in $P\left( t \right)={{P}_{0}}{{e}^{0.025t}}$:
$\begin{align}
& P\left( t \right)={{P}_{0}}{{e}^{0.025t}} \\
& P\left( 2 \right)=5000{{e}^{0.025\left( 2 \right)}} \\
& =5000{{e}^{0.05}}
\end{align}$
$\begin{align}
& P\left( 2 \right)=5000{{e}^{0.025}} \\
& \approx 5000\times 1.051271 \\
& \approx 5256.355 \\
& \approx 5256.36
\end{align}$
c)
Doubling time is the amount of time necessary for the investment to double in size.
If ${{P}_{0}}=\$5000$ is the investment at time 0, then the double of ${{P}_{0}}$ is :
$\begin{align}
& 2{{P}_{0}}=2\times \$5000\\&=\$10,000\end{align}$
Thus, to find the doubling time, replace $P\left( t \right)$ with $10,000$ and ${{P}_{0}}=\$5000$ in $P\left( t \right)={{P}_{0}}{{e}^{0.025t}}$ , substitute and solve for $t$:
$\begin{align}
& P\left( t \right)=5000{{e}^{0.025t}} \\
& 10,000=5000{{e}^{0.025t}} \\
\end{align}$
Divide by $5000$ on both the sides:
$\begin{align}
& \frac{10,000}{5000}=\frac{5000{{e}^{0.025t}}}{5000} \\
& 2={{e}^{0.025t}} \\
\end{align}$
Take the natural logarithm on both the sides:
$\ln 2=\ln {{e}^{0.025t}}$
Apply the power rule of logarithms $\ln {{m}^{n}}=n\ln m$ in $\ln {{e}^{0.025t}}$:
$\begin{align}
& \ln 2=\ln {{e}^{0.025t}} \\
& \ln 2=0.025t\operatorname{lne} \\
\end{align}$
$\begin{align}
& \ln 2=0.025t\ln e \\
& 0.6931=0.025t
\end{align}$
Divide by $0.025$ on both sides,
$\begin{align}
& \frac{0.6931}{0.025}=\frac{0.025t}{0.025} \\
& 27.7\approx t
\end{align}$
