1. **Problem statement:** We want to find how many years it will take for an investment of 50000 to grow to 60000 with an interest rate of 10% compounded quarterly. 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. **Given values:** - $P = 50000$ - $A = 60000$ - $r = 0.10$ - $n = 4$ (quarterly compounding) 4. **Substitute values into the formula:** $$60000 = 50000 \left(1 + \frac{0.10}{4}\right)^{4t}$$ 5. **Simplify inside the parentheses:** $$60000 = 50000 \left(1 + 0.025\right)^{4t} = 50000 (1.025)^{4t}$$ 6. **Divide both sides by 50000:** $$\frac{60000}{50000} = (1.025)^{4t}$$ $$1.2 = (1.025)^{4t}$$ 7. **Take the natural logarithm of both sides:** $$\ln(1.2) = \ln\left((1.025)^{4t}\right) = 4t \ln(1.025)$$ 8. **Solve for $t$:** $$t = \frac{\ln(1.2)}{4 \ln(1.025)}$$ 9. **Calculate the values:** $$\ln(1.2) \approx 0.182321556$$ $$\ln(1.025) \approx 0.02469261$$ 10. **Final calculation:** $$t = \frac{0.182321556}{4 \times 0.02469261} = \frac{0.182321556}{0.09877044} \approx 1.846$$ **Answer:** It will take approximately **1.85 years** for the investment to grow from 50000 to 60000 with 10% interest compounded quarterly.