Matthew invested \( \$ 5,300 \) in an account paying an interest rate of \( 2.1 \% \) compounded annually. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach \( \$ 6,020 \) ?

To find out how long it will take for Matthew's investment to grow from \$5,300 to \$6,020 at a compounded interest rate of 2.1%, we can use the formula for compound interest: \[ A = P(1 + r)^t \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial investment). - \( r \) is the annual interest rate (decimal). - \( t \) is the number of years the money is invested or borrowed. In this case: - \( A = 6,020 \) - \( P = 5,300 \) - \( r = 0.021 \) Plugging in the values: \[ 6,020 = 5,300(1 + 0.021)^t \] Dividing both sides by 5,300: \[ \frac{6,020}{5,300} = (1.021)^t \] Calculating the left side: \[ 1.13679 = (1.021)^t \] To solve for \( t \), we take the natural logarithm of both sides: \[ \ln(1.13679) = t \cdot \ln(1.021) \] Now we calculate \( t \): \[ t = \frac{\ln(1.13679)}{\ln(1.021)} \] Using a calculator: \[ t \approx \frac{0.1284}{0.0208} \approx 6.17 \text{ years} \] Rounding to the nearest year, it would take approximately **6 years** for the account to grow to \$6,020.