1. **State the problem:** A farmer stocked a pond with 1000 fish fingerlings. The fish population grows by a constant factor $x$ each year. After 2 years, the population is 36000. We need to find the annual growth rate $x$. 2. **Formula used:** The population growth can be modeled by the formula for exponential growth: $$ P_t = P_0 \times x^t $$ where $P_t$ is the population after $t$ years, $P_0$ is the initial population, and $x$ is the growth factor per year. 3. **Apply the known values:** Initial population $P_0 = 1000$ Population after 2 years $P_2 = 36000$ Time $t = 2$ 4. **Set up the equation:** $$ 36000 = 1000 \times x^2 $$ 5. **Solve for $x^2$:** $$ x^2 = \frac{36000}{1000} = 36 $$ 6. **Find $x$ by taking the square root:** $$ x = \sqrt{36} = 6 $$ 7. **Interpretation:** The annual growth factor is 6, meaning the fish population multiplies by 6 each year. 8. **Calculate the annual growth rate as a percentage:** The growth rate $r$ is given by: $$ r = (x - 1) \times 100\% = (6 - 1) \times 100\% = 500\% $$ **Final answer:** The annual growth rate of the fish population is 500%.