# Step-by-Step Solution ## 1. Identify the Variables Let's define: - Present Value (PV): $6,000 (the amount you have) - Future Value (FV): $9,000 (the amount you need) - Annual interest rate (r): 10% (.10 as a decimal) - Compounding periods per year (n): 12 (monthly) - Number of years (t): unknown (what we need to find) ## 2. Use the Compound Interest Formula The formula for future value with compounding is: \[ FV = PV \left(1 + \frac{r}{n}\right)^{nt} \] ## 3. Substitute the Values \[ 900 = 600 \left(1 + \frac{.10}{12}\right)^{12t} \] ## 4. Solve for \( t \) First, divide both sides by 600: \[ \frac{900}{600} = \left(1 + \frac{.10}{12}\right)^{12t} \] \[ 1.5 = \left(1 + \frac{.10}{12}\right)^{12t} \] Calculate \( 1 + \frac{.10}{12} \): \[ 1 + \frac{.10}{12} = 1 + .008333\ldots = 1.008333\ldots \] So, \[ 1.5 = (1.008333\ldots)^{12t} \] Now, take the natural logarithm of both sides: \[ \ln(1.5) = \ln\left((1.008333\ldots)^{12t}\right) \] \[ \ln(1.5) = 12t \cdot \ln(1.008333\ldots) \] \[ t = \frac{\ln(1.5)}{12 \cdot \ln(1.008333\ldots)} \] ## 5. Calculate the Value First, compute each logarithm: \[ \ln(1.5) \approx .405465 \] \[ \ln(1.008333\ldots) \approx .008298 \] Now, substitute: \[ t = \frac{.405465}{12 \times .008298} \] \[ 12 \times .008298 \approx .09958 \] \[ t \approx \frac{.405465}{.09958} \approx 4.075 \] ## Final Answer It will take approximately **4.08 years** for the $6,000 to grow to $9,000 at 10% interest compounded monthly.