Questions: Yolanda deposited 3000 into an account with a 3.5% annual interest rate, compounded quarterly. Assuming that no withdrawals are made, how long will it take for the investment to grow to 4500? Do not round any intermediate computations, and round your answer to the nearest hundredth.

Solution Steps

To find the time required for an investment to grow to a certain amount with compound interest, we can use the compound interest formula:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial amount of money).
• \( r \) is the annual interest rate (decimal).
• \( n \) is the number of times that interest is compounded per year.
• \( t \) is the time the money is invested for in years.

We need to solve for \( t \) given:

• \( A = 4500 \)
• \( P = 3000 \)
• \( r = 0.035 \)
• \( n = 4 \) (since the interest is compounded quarterly)

Rearrange the formula to solve for \( t \):

\[ t = \frac{\log\left(\frac{A}{P}\right)}{n \cdot \log\left(1 + \frac{r}{n}\right)} \]

We are given the following values:

• Principal amount, \( P = 3000 \)
• Future value, \( A = 4500 \)
• Annual interest rate, \( r = 0.035 \)
• Number of times interest is compounded per year, \( n = 4 \)

The compound interest formula is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

To solve for \( t \), we rearrange the formula: \[ t = \frac{\log\left(\frac{A}{P}\right)}{n \cdot \log\left(1 + \frac{r}{n}\right)} \]

Substitute the given values into the rearranged formula: \[ t = \frac{\log\left(\frac{4500}{3000}\right)}{4 \cdot \log\left(1 + \frac{0.035}{4}\right)} \]

Perform the calculations: \[ t = \frac{\log\left(1.5\right)}{4 \cdot \log\left(1 + 0.00875\right)} \] \[ t \approx \frac{0.1761}{4 \cdot 0.0087} \] \[ t \approx \frac{0.1761}{0.0348} \] \[ t \approx 5.0598 \]

Round the calculated value of \( t \) to the nearest hundredth: \[ t \approx 11.64 \]

Final Answer