\( \$ 1000 \) 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 \( \$ 2500 \) ? \[ \mathrm{t}=[?] \text { years } \] Round to the nearest year.

To solve this problem, we can use the formula for continuous compounding of interest: \[ A = P e^{rt} \] where: - \( A \) is the current amount in the account, - \( P \) is the initial deposit, - \( r \) is the interest rate, - \( t \) is the time in years. Given: - Initial deposit \( P = \$1000 \), - Interest rate \( r = 7\% = 0.07 \), - Current amount \( A = \$2500 \). We need to find the time \( t \) in years. Substitute the given values into the formula: \[ 2500 = 1000 e^{0.07t} \] Now, we can solve for \( t \) by rearranging the equation: \[ e^{0.07t} = \frac{2500}{1000} \] \[ e^{0.07t} = 2.5 \] Taking the natural logarithm of both sides: \[ 0.07t = \ln(2.5) \] \[ t = \frac{\ln(2.5)}{0.07} \] Now, we can calculate the value of \( t \) using a calculator. Calculate the value by following steps: - step0: Calculate: \(\frac{\ln{\left(2.5\right)}}{0.07}\) - step1: Convert the expressions: \(\frac{\ln{\left(\frac{5}{2}\right)}}{0.07}\) - step2: Rewrite the expression: \(\frac{\ln{\left(\frac{5}{2}\right)}}{\frac{7}{100}}\) - step3: Multiply by the reciprocal: \(\ln{\left(\frac{5}{2}\right)}\times \frac{100}{7}\) - step4: Multiply the terms: \(\frac{\ln{\left(\frac{5}{2}\right)}\times 100}{7}\) - step5: Multiply the terms: \(\frac{100\ln{\left(\frac{5}{2}\right)}}{7}\) The time \( t \) in years is approximately 13.09 years. Therefore, the \$1000 deposited into the account with a 7% interest rate, compounded continuously, was in the bank for approximately 13 years.