Questions: Lacey earned 6,231.00 from a summer job and put it in a savings account that earns 4% interest compounded continuously. When Lacey started college, she had 7,372.00 in the account which she used to pay for tuition. How long was the money in the account? Round your answer to the nearest month.

Solution Steps

To solve this problem, we need to use the formula for continuous compounding interest, which is given by:

\[ A = P e^{rt} \]

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).
• \( t \) is the time the money is invested for in years.
• \( e \) is the base of the natural logarithm.

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

• \( A = 7372 \)
• \( P = 6231 \)
• \( r = 0.04 \)

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

\[ t = \frac{\ln(A / P)}{r} \]

Finally, we will convert the time from years to months by multiplying by 12 and rounding to the nearest month.

We are given:

• Principal amount \( P = 6231.00 \)
• Final amount \( A = 7372.00 \)
• Annual interest rate \( r = 0.04 \)

The formula for continuous compounding interest is: \[ A = P e^{rt} \]

Rearranging the formula to solve for \( t \): \[ t = \frac{\ln\left(\frac{A}{P}\right)}{r} \]

Substitute \( A = 7372.00 \), \( P = 6231.00 \), and \( r = 0.04 \) into the formula: \[ t = \frac{\ln\left(\frac{7372.00}{6231.00}\right)}{0.04} \]

Using the given values: \[ t \approx \frac{\ln(1.183) }{0.04} \approx 4.2038 \text{ years} \]

Convert the time from years to months by multiplying by 12: \[ t_{\text{months}} = 4.2038 \times 12 \approx 50.4456 \]

Round the result to the nearest month: \[ t_{\text{months}} \approx 50 \]

Final Answer