\( \$ 600 \) were deposited into an account wit a \( 5 \% \) interest rate, compounded continuously. How many years was it in the bank if the current amount is \( \$ 6000 \) ? \[ t=[?] \text { years } \] 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 \( P = \$600 \), - Interest rate \( r = 5\% = 0.05 \), - Current amount \( A = \$6000 \). We need to find the time \( t \) in years. Substitute the given values into the formula: \[ 6000 = 600 e^{0.05t} \] Now, we can solve for \( t \) using the formula: \[ t = \frac{\ln\left(\frac{6000}{600}\right)}{0.05} \] Let's calculate the value of \( t \) using the formula. Calculate the value by following steps: - step0: Calculate: \(\frac{\ln{\left(\frac{6000}{600}\right)}}{0.05}\) - step1: Divide the terms: \(\frac{\ln{\left(10\right)}}{0.05}\) - step2: Rewrite the expression: \(\frac{\ln{\left(10\right)}}{\frac{1}{20}}\) - step3: Multiply by the reciprocal: \(\ln{\left(10\right)}\times 20\) - step4: Reorder the terms: \(20\ln{\left(10\right)}\) The time \( t \) in years is approximately 46.05 years. Therefore, the account was in the bank for approximately 46 years.