\( \$ 600 \) were deposited into an account with a \( 5 \% \) interest rate, compounded continuously. How many years was it in the bank if the current amount is \( \$ 6000 \) ? \( t=[?] \) 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 for continuous compounding of interest. Solve the equation by following steps: - step0: Solve for \(t\): \(6000=600e^{0.05t}\) - step1: Convert the expressions: \(6000=600e^{\frac{1}{20}t}\) - step2: Swap the sides: \(600e^{\frac{1}{20}t}=6000\) - step3: Divide both sides: \(\frac{600e^{\frac{1}{20}t}}{600}=\frac{6000}{600}\) - step4: Divide the numbers: \(e^{\frac{1}{20}t}=10\) - step5: Take the logarithm of both sides: \(\ln{\left(e^{\frac{1}{20}t}\right)}=\ln{\left(10\right)}\) - step6: Evaluate the logarithm: \(\frac{1}{20}t=\ln{\left(10\right)}\) - step7: Multiply by the reciprocal: \(\frac{1}{20}t\times 20=\ln{\left(10\right)}\times 20\) - step8: Multiply: \(t=20\ln{\left(10\right)}\) The solution to the equation is \( t = 20 \ln{(10)} \). Now, we can calculate the value of \( t \) by substituting the value of \( \ln{(10)} \) into the equation. \[ t = 20 \ln{(10)} \] \[ t \approx 20 \times 2.3026 \] \[ t \approx 46.052 \] Rounding to the nearest year, the time \( t \) is approximately 46 years.