1. **Problem:** Solve for $x$ in the equation $$8^{2x+1} = \frac{128^{x-2}}{4^x \times 16}$$ 2. **Step 1: Express all terms with base 2** - $8 = 2^3$ - $128 = 2^7$ - $4 = 2^2$ - $16 = 2^4$ 3. **Rewrite the equation:** $$\left(2^3\right)^{2x+1} = \frac{\left(2^7\right)^{x-2}}{2^{2x} \times 2^4}$$ 4. **Simplify exponents:** $$2^{3(2x+1)} = \frac{2^{7(x-2)}}{2^{2x+4}}$$ 5. **Simplify the right side:** $$2^{3(2x+1)} = 2^{7(x-2) - (2x+4)}$$ 6. **Expand exponents:** $$2^{6x+3} = 2^{7x - 14 - 2x - 4}$$ $$2^{6x+3} = 2^{5x - 18}$$ 7. **Since bases are equal, set exponents equal:** $$6x + 3 = 5x - 18$$ 8. **Solve for $x$:** $$6x - 5x = -18 - 3$$ $$x = -21$$ **Final answer:** $x = -21$