A principal of \( \$ 3800 \) is invested at \( 5.5 \% \) interest, compounded annually. How many years will it take to accumulate \( \$ 8000 \) or more in the account? (Use the calculator provided if necessary.) Write the smallest possible whole number answer.

To find out how long it will take for an investment of \( \$3800 \) to grow to \( \$8000 \) at an annual interest rate of \( 5.5\% \), we can use the formula for compound interest, which is: \[ A = P(1 + r)^t \] where \( A \) is the amount of money accumulated after n years, including interest, \( P \) is the principal amount (the initial amount), \( r \) is the annual interest rate (decimal), and \( t \) is the number of years the money is invested or borrowed. Setting \( A = 8000 \), \( P = 3800 \), and \( r = 0.055 \): \[ 8000 = 3800(1 + 0.055)^t \] Dividing both sides by \( 3800 \): \[ \frac{8000}{3800} = (1.055)^t \] Calculating the left side gives: \[ 2.10526 \approx (1.055)^t \] Next, we take the natural logarithm of both sides: \[ \ln(2.10526) = t \cdot \ln(1.055) \] Now solve for \( t \): \[ t = \frac{\ln(2.10526)}{\ln(1.055)} \approx \frac{0.7411}{0.0535} \approx 13.87 \] Since we want the smallest whole number, we round up to get \( 14 \) years. So the answer is \( 14 \).