1. **State the problem:** Solve the inequality $$49 \cdot \left(\frac{1}{8}\right)^{-2x-6} > \left(\frac{1}{32}\right)^{-x+11}$$. 2. **Rewrite bases as powers of primes:** - $49 = 7^2$ - $\frac{1}{8} = 8^{-1} = 2^{-3}$ - $\frac{1}{32} = 32^{-1} = 2^{-5}$ 3. **Rewrite the inequality using these bases:** $$7^2 \cdot \left(2^{-3}\right)^{-2x-6} > \left(2^{-5}\right)^{-x+11}$$ 4. **Simplify exponents:** $$7^2 \cdot 2^{3(2x+6)} > 2^{5(x-11)}$$ 5. **Rewrite:** $$7^2 \cdot 2^{6x+18} > 2^{5x - 55}$$ 6. **Divide both sides by $2^{5x - 55}$:** $$7^2 \cdot \frac{2^{6x+18}}{2^{5x - 55}} > 1$$ 7. **Use exponent subtraction:** $$7^2 \cdot 2^{(6x+18) - (5x - 55)} > 1$$ 8. **Simplify exponent:** $$7^2 \cdot 2^{x + 73} > 1$$ 9. **Rewrite $7^2$ as $49$:** $$49 \cdot 2^{x + 73} > 1$$ 10. **Divide both sides by 49:** $$\cancel{49} \cdot 2^{x + 73} > \frac{1}{\cancel{49}}$$ $$2^{x + 73} > \frac{1}{49}$$ 11. **Rewrite $\frac{1}{49} = 49^{-1} = 7^{-2}$, but better to take log base 2:** 12. **Take logarithm base 2 of both sides:** $$x + 73 > \log_2\left(\frac{1}{49}\right) = -\log_2(49)$$ 13. **Calculate $\log_2(49)$:** $$49 = 7^2 \Rightarrow \log_2(49) = 2 \log_2(7)$$ 14. **Final inequality:** $$x + 73 > -2 \log_2(7)$$ 15. **Solve for $x$:** $$x > -73 - 2 \log_2(7)$$ --- **Final answer:** $$\boxed{x > -73 - 2 \log_2(7)}$$