Mackenzie invested \( \$ 770 \) in an account paying an interest rate of \( 6.1 \% \) compounded continuously. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach \( \$ 1,830 \) ?

To determine how long it will take for Mackenzie's investment to grow from \$770 to \$1,830 with continuous compounding at an annual interest rate of 6.1%, we can use the formula for continuous compound interest: \[ A = P e^{rt} \] Where: - \( A \) is the amount of money accumulated after time \( t \). - \( P \) is the principal amount (\$770). - \( r \) is the annual interest rate (6.1% or 0.061 in decimal). - \( t \) is the time in years. - \( e \) is the base of the natural logarithm. **Step-by-Step Calculation:** 1. **Set Up the Equation:** \[ 1830 = 770 \times e^{0.061t} \] 2. **Divide Both Sides by 770:** \[ \frac{1830}{770} = e^{0.061t} \] \[ 2.376623 = e^{0.061t} \] 3. **Take the Natural Logarithm of Both Sides:** \[ \ln(2.376623) = 0.061t \] \[ t = \frac{\ln(2.376623)}{0.061} \] 4. **Calculate the Natural Logarithm:** \[ \ln(2.376623) \approx 0.8657 \] 5. **Solve for \( t \):** \[ t = \frac{0.8657}{0.061} \approx 14.2 \text{ years} \] 6. **Round to the Nearest Year:** \[ t \approx 14 \text{ years} \] **Conclusion:** It will take approximately **14 years** for Mackenzie's investment to grow to \$1,830 under the given conditions. **Answer:** Approximately 14 years are needed for the account to reach \$1,830.