To solve this we are going to use the continuous compounding interest formula:
[tex]A=Pe^{rt}[/tex]
where:
[tex]A[/tex] is the amount after [tex]t[/tex] years.
[tex]P[/tex] is the initial investment.
[tex]e[/tex] is the Euler's constant.
[tex]r[/tex] is the interest rate.
Since we know from our problem that the investment of 3000 doubles in value in 8.3 years, [tex]A=2(3000)=6000[/tex]. We also know that [tex]P=3000[/tex] and [tex]t=8.3[/tex], so lets replace those values in our equation:
[tex]A=Pe^{8.3r}[/tex]
[tex]6000=3000e^{rt}[/tex]
Notice that thing we are trying to find, [tex]r[/tex], is in the exponent; therefore, we must use logarithms to bring it down:
[tex] \frac{6000}{3000} =e^{8.3r}[/tex]
[tex]2=e^{8.3r}[/tex]
[tex]e^{8.3r}=2[/tex]
[tex]ln(e^{8.3r})=ln(2)[/tex]
[tex]8.3r=ln(2)[/tex]
[tex]r= \frac{ln(2)}{8.3} [/tex]
[tex]r=0.084[/tex]
The only thing left now is multiplying our rate by 100% to express it as a percentage:
[tex](0.084)(100)=8.4[/tex]%
We can conclude that the continuous compounding interest rate needed to double $3000 in 8.3 years is approximately 8.4%.
