Noah is going to invest in an account paying an interest rate of \( 7 \% \) compounded continuously. How much would Noah need to invest, to the nearest hundred dollars, for the value of the account to reach \( \$ 12,900 \) in 10 years?

To solve this problem, we can use the formula for continuous compounding, which is given by: \[ A = Pe^{rt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( t \) is the time the money is invested for, in years. - \( e \) is the base of the natural logarithm (approximately equal to 2.71828). We know: - \( A = 12,900 \) - \( r = 0.07 \) - \( t = 10 \) We need to solve for \( P \): \[ 12,900 = P e^{0.07 \times 10} \] Calculating \( e^{0.7} \): \[ e^{0.7} \approx 2.01375 \] Now, substituting this back into the equation: \[ 12,900 = P \times 2.01375 \] To find \( P \): \[ P = \frac{12,900}{2.01375} \approx 6,397.52 \] Rounding to the nearest hundred dollars, Noah would need to invest approximately **$6,400**.