1. **State the problem:** Simplify the expression $$\sqrt{\frac{25w^6}{x^2}} \div wx^3$$ leaving the answer with positive exponents. 2. **Rewrite the expression:** The division can be written as multiplication by the reciprocal: $$\sqrt{\frac{25w^6}{x^2}} \times \frac{1}{wx^3}$$ 3. **Simplify inside the square root:** $$\sqrt{\frac{25w^6}{x^2}} = \frac{\sqrt{25w^6}}{\sqrt{x^2}}$$ 4. **Evaluate the square roots:** $$\sqrt{25w^6} = 5w^3$$ because $$\sqrt{w^6} = w^{6/2} = w^3$$ $$\sqrt{x^2} = x$$ So, $$\sqrt{\frac{25w^6}{x^2}} = \frac{5w^3}{x}$$ 5. **Substitute back:** $$\frac{5w^3}{x} \times \frac{1}{wx^3} = \frac{5w^3}{x} \times \frac{1}{wx^3}$$ 6. **Multiply the fractions:** $$\frac{5w^3 \times 1}{x \times wx^3} = \frac{5w^3}{w x^{4}}$$ 7. **Cancel common factors:** $$\frac{5\cancel{w^3}}{\cancel{w} x^{4}} = \frac{5w^{3-1}}{x^{4}} = \frac{5w^{2}}{x^{4}}$$ 8. **Final answer:** $$\boxed{\frac{5w^{2}}{x^{4}}}$$ This expression has only positive exponents as required.