Suppose that \( \$ 2000 \) is invested at an interest rate of \( 2.5 \% \) per year, compounded continuously. After how many years will the initial investment be doubled?

To find out how many years it will take for the initial investment of \$2000 to double at a continuously compounded interest rate of 2.5%, we can use the formula for continuous compounding, which is given by: \[ A = Pe^{rt} \] Where: - \( A \) is the amount of money accumulated after time \( t \), - \( P \) is the principal amount (the initial investment), - \( r \) is the annual interest rate (as a 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 want to double the initial investment, so: \[ A = 2P \] Plugging in the values we have: \[ 2P = Pe^{rt} \] Dividing both sides by \( P \) (assuming \( P \neq 0 \)): \[ 2 = e^{rt} \] Now substituting \( r = 0.025 \): \[ 2 = e^{0.025t} \] To solve for \( t \), we take the natural logarithm of both sides: \[ \ln(2) = 0.025t \] Now we can solve for \( t \): \[ t = \frac{\ln(2)}{0.025} \] Calculating \( t \): \[ t \approx \frac{0.693147}{0.025} \approx 27.72588 \] Rounding to two decimal places, it will take approximately \( 27.73 \) years for the initial investment to double.