Keisha deposited \( \$ 2500 \) into an account with a \( 5.9 \% \) annual interest rate, compounded semiannually. Assuming no other deposits or withdrawals are made, how long will it take for the investment to grow to \( \$ 4000 \) ? Round to the nearest hundredth.

To solve this problem, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial deposit or investment). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years. In this case: - \( A = 4000 \) - \( P = 2500 \) - \( r = 0.059 \) - \( n = 2 \) (since it's compounded semiannually) Now plug in the values: \[ 4000 = 2500 \left(1 + \frac{0.059}{2}\right)^{2t} \] This simplifies to: \[ 4000 = 2500 \left(1 + 0.0295\right)^{2t} \] Calculating \(1 + 0.0295 = 1.0295\): \[ 4000 = 2500 (1.0295)^{2t} \] Next, divide both sides by \(2500\): \[ \frac{4000}{2500} = (1.0295)^{2t} \] This simplifies to: \[ 1.6 = (1.0295)^{2t} \] Now we can take the natural logarithm of both sides: \[ \ln(1.6) = 2t \cdot \ln(1.0295) \] Solving for \(t\): \[ t = \frac{\ln(1.6)}{2 \cdot \ln(1.0295)} \] Calculating the natural logs: \[ t \approx \frac{0.4700}{2 \cdot 0.0292} \approx \frac{0.4700}{0.0584} \approx 8.04 \] Rounded to the nearest hundredth, it will take approximately **8.04 years** for Keisha's investment to grow to $4000.