Investment Principal 012864

1. **State the problem:** David wants to know how much to invest now (the principal $P$) to have $63,400$ after $12$ years with an interest rate of $2.24\%$ compounded monthly. 2. **Formula used:** The compound interest formula is: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after time $t$ - $P$ is the principal (initial investment) - $r$ is the annual interest rate (decimal) - $n$ is the number of compounding periods per year - $t$ is the time in years 3. **Identify values:** - $A = 63400$ - $r = 2.24\% = 0.0224$ - $n = 12$ (monthly compounding) - $t = 12$ 4. **Rearrange formula to solve for $P$:** $$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}$$ 5. **Calculate the denominator:** $$1 + \frac{0.0224}{12} = 1 + 0.0018667 = 1.0018667$$ 6. **Calculate the exponent:** $$nt = 12 \times 12 = 144$$ 7. **Calculate the compound factor:** $$\left(1.0018667\right)^{144}$$ 8. **Calculate $P$:** $$P = \frac{63400}{\left(1.0018667\right)^{144}}$$ 9. **Intermediate step with cancellation:** $$P = \frac{63400}{\cancel{\left(1.0018667\right)^{144}}} \times \cancel{\left(1.0018667\right)^{-144}} = 63400 \times \left(1.0018667\right)^{-144}$$ 10. **Evaluate $\left(1.0018667\right)^{144}$:** $$\left(1.0018667\right)^{144} \approx 1.308$$ 11. **Calculate $P$:** $$P = \frac{63400}{1.308} \approx 48472.58$$ **Final answer:** David needs to invest approximately **48472.58** now to have $63,400$ after 12 years with $2.24\%$ interest compounded monthly.