Jennifer deposited \( \$ 3200 \) into an account earning \( 4.2 \% \), compounded monthly. Assuming no withdrawals or deposits are made, how long will it take for the investment to grow to \( \$ 10,000 \). Rounded to the nearest hundredth.

To find out how long it will take for Jennifer's investment to grow from \( \$ 3200 \) to \( \$ 10,000 \) at an interest rate of \( 4.2\% \) compounded monthly, 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 amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the number of years the money is invested for. Plugging in the values, we have: - \( A = 10000 \) - \( P = 3200 \) - \( r = 0.042 \) - \( n = 12 \) Substituting these into the formula gives us: \[ 10000 = 3200 \left(1 + \frac{0.042}{12}\right)^{12t} \] Simplifying the calculation: 1. Calculate \( 1 + \frac{0.042}{12} = 1 + 0.0035 = 1.0035 \). 2. Divide both sides by \( 3200 \): \[ \frac{10000}{3200} = \left(1.0035\right)^{12t} \] \[ 3.125 = 1.0035^{12t} \] 3. Now, take the logarithm of both sides: \[ \log(3.125) = 12t \cdot \log(1.0035) \] 4. Solve for \( t \): \[ t = \frac{\log(3.125)}{12 \cdot \log(1.0035)} \] 5. Calculate the logarithms: Using a calculator: - \( \log(3.125) \approx 0.4948 \) - \( \log(1.0035) \approx 0.00152 \) Substitute these values: \[ t \approx \frac{0.4948}{12 \cdot 0.00152} \approx \frac{0.4948}{0.01824} \approx 27.16 \] So, it will take approximately **27.16 years** for Jennifer's investment to grow to \( \$ 10,000 \).