Nathan is going to invest \( \$ 80,000 \) and leave it in an account for 16 years. Assuming the interest is compounded continuously, what interest rate, to the nearest hundredth of a percent, would be required in order for Nathan to end up with \( \$ 160,000 \) ?

To find the interest rate required for Nathan to double his investment through continuous compounding, we can use the formula: \[ A = Pe^{rt} \] where: - \( A \) is the amount of money accumulated after time \( t \), - \( P \) is the principal amount (the initial amount of money), - \( r \) 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 equal to 2.71828). Given: - \( A = 160,000 \) - \( P = 80,000 \) - \( t = 16 \) We can substitute these values into the formula: \[ 160,000 = 80,000 \cdot e^{16r} \] To solve for \( r \), we first divide both sides by \( 80,000 \): \[ 2 = e^{16r} \] Next, we take the natural logarithm of both sides: \[ \ln(2) = 16r \] Now, we can solve for \( r \): \[ r = \frac{\ln(2)}{16} \] Using a calculator to find \( \ln(2) \), which is approximately \( 0.693147 \): \[ r = \frac{0.693147}{16} \approx 0.0433 \] To express \( r \) as a percentage, we multiply by 100: \[ r \approx 4.33\% \] Thus, the interest rate required for Nathan to double his investment in 16 years with continuous compounding is approximately **4.33%**.