1. We are given the formula for compound interest: $A = P(1 + i)^n$. Here, $A$ is the amount after interest, $P$ is the principal, $i$ is the interest rate per period, and $n$ is the number of periods. 2. In the problem, the values are: $A = 19319.48$, $P = 18500$, and $n = 6$. 3. The formula is used as: $$19319.48 = 18500 \times (1 + \frac{r}{1200})^6$$ where $r$ is the annual interest rate as a percentage and the monthly interest rate is $\frac{r}{1200}$ since the year is divided into 12 months and rate into decimals. 4. To isolate $(1 + \frac{r}{1200})$, divide both sides by 18500: $$\frac{19319.48}{18500} = (1 + \frac{r}{1200})^6$$ $$1.04429 \approx (1 + \frac{r}{1200})^6$$ 5. To solve for $1 + \frac{r}{1200}$, take the 6th root of both sides. The 6th root symbol shown in the image is the root notation for: $$1 + \frac{r}{1200} = \sqrt[6]{1.04429}$$ 6. Calculating the 6th root means finding the number that, when raised to the power of 6, equals 1.04429. This is why the square root symbol with the index 6 (the 6th root) is used. 7. Evaluating the 6th root: $$1 + \frac{r}{1200} \approx 1.00725$$ 8. Subtract 1 to find $\frac{r}{1200}$: $$\frac{r}{1200} = 0.00725$$ 9. Multiply both sides by 1200 to isolate $r$: $$r = 0.00725 \times 1200 = 8.7\%$$ 10. Thus, the annual interest rate $r$ is approximately 8.7%. Summary: The image shows taking the 6th root because in the formula the term $(1 + r/1200)$ is raised to the 6th power. To solve for $(1 + r/1200)$, we must undo the exponentiation by using the 6th root, represented by the radical with a small 6 above it.