\( \$ 1000 \) were deposited into an account with a \( 7 \% \) interest rate, compounded continuously. How many years was it in the bank if the current amount is \( \$ 2500 \) ? \[ \mathrm{t}=[\text { ? ] years } \]

To determine how many years the money was in the bank with continuous compounding, we can use the formula for continuous compound interest: \[ A = P \cdot e^{rt} \] Where: - \( A \) is the amount of money accumulated after \( t \) years, including interest. - \( P \) is the principal amount (initial deposit). - \( r \) is the annual interest rate (in decimal). - \( t \) is the time the money is invested for, in years. - \( e \) is the base of the natural logarithm, approximately equal to 2.71828. Given: - \( A = \$2500 \) - \( P = \$1000 \) - \( r = 7\% = 0.07 \) Plugging these values into the formula: \[ 2500 = 1000 \cdot e^{0.07t} \] 1. **Divide both sides by 1000:** \[ 2.5 = e^{0.07t} \] 2. **Take the natural logarithm of both sides to solve for \( t \):** \[ \ln(2.5) = 0.07t \] 3. **Solve for \( t \):** \[ t = \frac{\ln(2.5)}{0.07} \] 4. **Calculate the value:** \[ \ln(2.5) \approx 0.9163 \] \[ t \approx \frac{0.9163}{0.07} \approx 13.09 \text{ years} \] **Answer:** Approximately 13.09 years were required for the deposit to grow to \$2500.