Step 1 :Given values are: Future value of the investment (S) = $40,000, Initial investment (P) = $10,000, Annual interest rate (r) = 12%, Number of times interest is compounded per year (n) = 12.
Step 2 :We need to find the time (t) it takes for the investment to grow to $40,000.
Step 3 :We use the formula for the future value of an investment compounded monthly: \(S = P \left(1 + \frac{r}{n}\right)^{nt}\).
Step 4 :Rearranging the formula to solve for t, we get \(t = \frac{\ln(\frac{S}{P})}{n \ln(1 + \frac{r}{n})}\).
Step 5 :Substituting the given values into the formula, we get \(t = \frac{\ln(\frac{40000}{10000})}{12 \ln(1 + \frac{0.12}{12})}\).
Step 6 :Calculating the above expression, we find that \(t \approx 11.61\).
Step 7 :So, it will take approximately 11.61 years for the investment to grow to $40,000.
Step 8 :Final Answer: The investment will grow to $40,000 in \(\boxed{11.61}\) years.