Question

Julia invested $17,000 in an account paying an interest rate of 2.4% compounded daily. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $21,400?

Answer 1

To solve this problem, we must use the formula for Compound Interest:

$P_N=P_0\cdot(1+\frac{r}{k})^{N\cdot k}.$

Where:

• P_N is the balance in the account after N years,

,

• P_0 is the starting balance of the account (also called an initial deposit, or principal),

,

• r is the annual interest rate in decimal form,

,

• k is the number of compounding periods in one year.

In this problem, we have:

• P_0 = $17,000,

,

• P_N = $21,400,

,

• r = 2.4% = 0.024,

,

• k = 365 (the interest rate is compounded daily).

Replacing the data of the problem in the equation above, we have:

$21400=17000\cdot(1+\frac{0.024}{365})^{N\cdot365}\text{.}$

Solving for N the last equation, we get:

$\begin{gathered} \frac{21400}{17000}=(1+\frac{0.024}{365})^{N\cdot365}, \\ \ln (\frac{21400}{17000})=N\cdot365\cdot\ln (1+\frac{0.024}{365}), \\ N=\frac{\ln(\frac{21400}{17000})}{365\cdot\ln(1+\frac{0.024}{365})}\cong9.6. \end{gathered}$

Answer

To the nearest tenth of a year, it will take 9.6 years for the account to reach $21,400.