1. **State the problem:** Rewrite the expression $$\frac{(9^{-3} \cdot 2^{-7})^2}{9^5}$$ using only positive exponents and combine powers with the same base. 2. **Recall exponent rules:** - Power of a product: $$(a \cdot b)^m = a^m \cdot b^m$$ - Power of a power: $$(a^m)^n = a^{m \cdot n}$$ - Division of same bases: $$\frac{a^m}{a^n} = a^{m-n}$$ - Negative exponent: $$a^{-m} = \frac{1}{a^m}$$ 3. **Apply power of a product:** $$ (9^{-3} \cdot 2^{-7})^2 = 9^{-3 \cdot 2} \cdot 2^{-7 \cdot 2} = 9^{-6} \cdot 2^{-14} $$ 4. **Rewrite the original expression:** $$ \frac{9^{-6} \cdot 2^{-14}}{9^5} $$ 5. **Divide powers with the same base 9:** $$ 9^{-6} \div 9^5 = 9^{-6 - 5} = 9^{-11} $$ 6. **Rewrite the expression:** $$ 9^{-11} \cdot 2^{-14} $$ 7. **Rewrite with positive exponents:** $$ 9^{-11} = \frac{1}{9^{11}}, \quad 2^{-14} = \frac{1}{2^{14}} $$ 8. **Final expression:** $$ \frac{1}{9^{11} \cdot 2^{14}} $$ **Answer:** $$\boxed{\frac{1}{9^{11} \cdot 2^{14}}}$$