Question

At what interest compounded continuously must you invest $4900 to have$10740.65 money in 9 years? First round the interest rate r to 4 decimalplaces and then re-write it as aa percentage with 2 decimal places.

Answer 1

ANSWER

$\begin{gathered} r=0.0872 \\ r=8.72\% \end{gathered}$

EXPLANATION

We want to find the rate at which the amount was continuously compounded.

The formula for the total amount for a continuously compounded principal is:

$A=Pe^{rt}$

where A = amount

P = principal

r = rate

t = time (in years)

Substituting the given values into the equation:

$\begin{gathered} 10740.65=4900\cdot e^{r\cdot9} \\ 10740.65=4900\cdot e^{9r} \end{gathered}$

Divide both sides by 4900:

$\begin{gathered} \frac{10740.65}{4900}=e^{9r} \\ e^{9r}=2.1920 \end{gathered}$

Find the natural logarithm of both sides of the equation:

$\begin{gathered} \ln (e^{9r})=\ln 2.1920 \\ 9r=0.7848 \end{gathered}$

Divide both sides by 9:

$\begin{gathered} r=\frac{0.7848}{9} \\ r=0.0872 \end{gathered}$

Convert to decimal number:

$\begin{gathered} r=0.0872\cdot100 \\ r=8.72\% \end{gathered}$

That is the interest rate.