Carter invested \( \$ 16,000 \) in an account paying an interest rate of \( 5.6 \% \) compounded monthly. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach \( \$ 29,600 \) ?

To determine how long it takes for Carter's investment of \( \$ 16,000 \) to grow to \( \$ 29,600 \) with an interest rate of \( 5.6\% \) compounded monthly, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] 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 in years. In this case: - \( A = 29,600 \) - \( P = 16,000 \) - \( r = 0.056 \) - \( n = 12 \) (since the interest is compounded monthly) Plugging in these values, we get: \[ 29,600 = 16,000 \left(1 + \frac{0.056}{12}\right)^{12t} \] First, divide both sides by \( 16,000 \): \[ 1.85 = \left(1 + \frac{0.056}{12}\right)^{12t} \] Calculating \( 1 + \frac{0.056}{12} \): \[ 1 + \frac{0.056}{12} \approx 1 + 0.00466667 \approx 1.00466667 \] Now we have: \[ 1.85 = (1.00466667)^{12t} \] To solve for \( t \), we take the natural logarithm of both sides: \[ \ln(1.85) = 12t \cdot \ln(1.00466667) \] Now we can isolate \( t \): \[ t = \frac{\ln(1.85)}{12 \cdot \ln(1.00466667)} \] Calculating the logarithms: 1. \( \ln(1.85) \approx 0.620576 \) 2. \( \ln(1.00466667) \approx 0.004654 \) So, \[ t = \frac{0.620576}{12 \cdot 0.004654} \approx \frac{0.620576}{0.055848} \approx 11.09 \] Rounding to the nearest year, it would take approximately \( 11 \) years for the account to reach \( \$29,600 \).