Question

An initial investment of 3000 doubles in value in 8.3 years. Assuming continuous compounding, what was the interest rate?

Answer 1

To solve this we are going to use the continuous compounding interest formula:
$A=Pe^{rt}$
where:
$A$ is the amount after $t$ years.
$P$ is the initial investment.
$e$ is the Euler's constant.
$r$ is the interest rate.

Since we know from our problem that the investment of 3000 doubles in value in 8.3 years, $A=2(3000)=6000$. We also know that $P=3000$ and $t=8.3$, so lets replace those values in our equation:
$A=Pe^{8.3r}$
$6000=3000e^{rt}$
Notice that thing we are trying to find, $r$, is in the exponent; therefore, we must use logarithms to bring it down:
$\frac{6000}{3000} =e^{8.3r}$
$2=e^{8.3r}$
$e^{8.3r}=2$
$ln(e^{8.3r})=ln(2)$
$8.3r=ln(2)$
$r= \frac{ln(2)}{8.3}$
$r=0.084$
The only thing left now is multiplying our rate by 100% to express it as a percentage:
$(0.084)(100)=8.4$%

We can conclude that the continuous compounding interest rate needed to double $3000 in 8.3 years is approximately 8.4%.