How much would you have to deposit in an account with a \( 7.5 \% \) interest rate, compounded continuously, to have \( \$ 2500 \) in your account 8 years later? \[ P=\$[?] \]

To determine how much you need to deposit now (\( P \)) to have \$2,500 in your account after 8 years with a continuous compounding interest rate of \( 7.5\% \), you can use the formula for continuous compounding: \[ A = P e^{rt} \] Where: - \( A \) is the amount of money accumulated after \( t \) years, including interest. - \( P \) is the principal amount (the initial deposit). - \( r \) is the annual interest rate (in decimal). - \( t \) is the time the money is invested for in years. - \( e \) is the base of the natural logarithm (approximately 2.71828). Given: - \( A = \$2,500 \) - \( r = 7.5\% = 0.075 \) - \( t = 8 \) years **Step 1: Rearrange the formula to solve for \( P \)** \[ P = \frac{A}{e^{rt}} \] **Step 2: Plug in the known values** \[ P = \frac{2500}{e^{0.075 \times 8}} = \frac{2500}{e^{0.6}} \] **Step 3: Calculate \( e^{0.6} \)** \[ e^{0.6} \approx 1.8221188 \] **Step 4: Solve for \( P \)** \[ P = \frac{2500}{1.8221188} \approx 1371.46 \] **Final Answer:** \[ P = \$1,\!371.46 \]