Question

Molly invests $8,700 into her son‘s college fund, which earns 2% annual interest compounded daily. Find when the value of the fund reaches $12,000.

Answer 1

To answer this question we will use the following formula for compounded daily interest:

$\begin{gathered} A=A_0(1+\frac{r}{365})^{365t},_{} \\ \text{where A}_0\text{ is the initial amount, r is the annual rate as a decimal number, } \\ \text{and t is the number of years.} \end{gathered}$

Substituting A₀=8700, A=12000, and r=0.02 we get:

$12000=8700(1+\frac{0.02}{365})^{365t}\text{.}$

Dividing the above result by 8700 we get:

$\begin{gathered} \frac{12000}{8700}=\frac{8700}{8700}(1+\frac{0.02}{365})^{365t}, \\ \frac{40}{29}=(1+\frac{0.02}{365})^{365t}\text{.} \end{gathered}$

Applying the natural logarithm we get:

$\ln (\frac{40}{29})=365t\cdot\ln (1+\frac{0.02}{365})\text{.}$

Finally, dividing the above equation by

$365\cdot\ln (1+\frac{0.02}{365})$

we get:

$t=\frac{\ln(\frac{40}{29})}{365\ln(1+\frac{0.02}{365})}\text{.}$

Therefore:

$t\approx16.08$

Answer: The value of the fund reaches $12,000 after 16.08 years.