\( \$ 1000 \) are deposited in an account with a \( 8.5 \% \) interest rate, compounded continuously. What is the balance after 5 years? Begin by entering \( P \), or the principal (initial investment). \( P=\$[?] \)

To find the balance after 5 years with continuous compounding, we can use the formula for continuous compounding: \[ A = P \times e^{rt} \] where: - \( A \) is the balance after \( t \) years, - \( P \) is the principal (initial investment), - \( r \) is the interest rate, - \( t \) is the time in years, - \( e \) is the base of the natural logarithm. Given: - Principal (initial investment) \( P = \$1000 \), - Interest rate \( r = 8.5\% = 0.085 \), - Time \( t = 5 \) years. Substitute the values into the formula: \[ A = 1000 \times e^{0.085 \times 5} \] Now, we can calculate the balance after 5 years. Calculate the value by following steps: - step0: Calculate: \(1000e^{0.085\times 5}\) - step1: Multiply the numbers: \(1000e^{0.425}\) - step2: Convert the expressions: \(1000e^{\frac{17}{40}}\) - step3: Rewrite the expression: \(1000\sqrt[40]{e^{17}}\) The balance after 5 years with continuous compounding is approximately $1529.59.