1. **State the problem:** We want to find the number of periods $n$ such that the future value (FV) of an investment grows from a present value (PV) of 15000 to 18640.84 with an interest rate $i=0.035$ per period. 2. **Formula used:** The compound interest formula is $$FV = PV (1 + i)^n$$ where $FV$ is the future value, $PV$ is the present value, $i$ is the interest rate per period, and $n$ is the number of periods. 3. **Substitute known values:** $$18640.84 = 15000 (1.035)^n$$ 4. **Isolate the exponential term:** $$\frac{18640.84}{15000} = (1.035)^n$$ $$1.2427227 = (1.035)^n$$ 5. **Take the natural logarithm of both sides:** $$\ln(1.2427227) = \ln((1.035)^n)$$ $$\ln(1.2427227) = n \ln(1.035)$$ 6. **Solve for $n$:** $$n = \frac{\ln(1.2427227)}{\ln(1.035)}$$ 7. **Calculate the values:** $$\ln(1.2427227) \approx 0.2179$$ $$\ln(1.035) \approx 0.0344$$ $$n = \frac{0.2179}{0.0344} \approx 6.3167$$ **Final answer:** The number of periods $n$ is approximately **6.3167**. This means it takes a little over 6 periods for the investment to grow from 15000 to 18640.84 at 3.5% interest per period.