\( \$ 800 \) were deposited into an account wit a \( 9.5 \% \) interest rate, compounded continuously. How many years was it in the bank if the current amount is \( \$ 6400 \) ? \[ t=[?] \text { years } \]

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 \( P = \$800 \), - Interest rate \( r = 9.5\% = 0.095 \), - Current amount \( A = \$6400 \). We need to find the time \( t \) in years. Substitute the given values into the formula: \[ 6400 = 800 e^{0.095t} \] Now, we can solve for \( t \) using the formula for continuous compounding of interest. Solve the equation by following steps: - step0: Solve for \(t\): \(6400=800e^{0.095t}\) - step1: Convert the expressions: \(6400=800e^{\frac{19}{200}t}\) - step2: Swap the sides: \(800e^{\frac{19}{200}t}=6400\) - step3: Divide both sides: \(\frac{800e^{\frac{19}{200}t}}{800}=\frac{6400}{800}\) - step4: Divide the numbers: \(e^{\frac{19}{200}t}=8\) - step5: Take the logarithm of both sides: \(\ln{\left(e^{\frac{19}{200}t}\right)}=\ln{\left(8\right)}\) - step6: Evaluate the logarithm: \(\frac{19}{200}t=\ln{\left(8\right)}\) - step7: Multiply by the reciprocal: \(\frac{19}{200}t\times \frac{200}{19}=\ln{\left(8\right)}\times \frac{200}{19}\) - step8: Multiply: \(t=\frac{200\ln{\left(8\right)}}{19}\) - step9: Simplify: \(t=\frac{600\ln{\left(2\right)}}{19}\) The time \( t \) in years is given by the formula: \[ t = \frac{600 \ln(2)}{19} \] Substitute the value of \( \ln(2) \) to find the numerical value of \( t \). Calculate the value by following steps: - step0: Calculate: \(\frac{600\ln{\left(2\right)}}{19}\) The time \( t \) in years is approximately 21.89 years.