Investment Growth Time 72857D

1. **State the problem:** We want to find the time $t$ it takes for an investment of $23750$ to grow to $27000$ at an interest rate of $1.59\%$ compounded semi-annually. 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:** - $A = 27000$ - $P = 23750$ - $r = 0.0159$ (since $1.59\% = 0.0159$) - $n = 2$ (semi-annually means twice a year) 4. **Plug values into formula:** $$27000 = 23750 \left(1 + \frac{0.0159}{2}\right)^{2t}$$ 5. **Simplify inside the parentheses:** $$1 + \frac{0.0159}{2} = 1 + 0.00795 = 1.00795$$ 6. **Divide both sides by 23750:** $$\frac{27000}{23750} = \left(1.00795\right)^{2t}$$ 7. **Calculate the left side:** $$1.136842105 = \left(1.00795\right)^{2t}$$ 8. **Take the natural logarithm of both sides:** $$\ln(1.136842105) = \ln\left(1.00795^{2t}\right)$$ 9. **Use log power rule:** $$\ln(1.136842105) = 2t \ln(1.00795)$$ 10. **Solve for $t$:** $$t = \frac{\ln(1.136842105)}{2 \ln(1.00795)}$$ 11. **Calculate logarithms:** $$\ln(1.136842105) \approx 0.1282$$ $$\ln(1.00795) \approx 0.00792$$ 12. **Calculate $t$:** $$t = \frac{0.1282}{2 \times 0.00792} = \frac{0.1282}{0.01584} \approx 8.09$$ 13. **Interpretation:** It takes approximately $8.09$ years for the investment to grow from $23750$ to $27000$ at $1.59\%$ interest compounded semi-annually. **Final answer:** $$\boxed{8.09 \text{ years}}$$