1. **State the problem:** Solve the equation $$16^x \times \left(\frac{1}{2}\right)^x = 4^{x+3}$$ for $x$. 2. **Rewrite the bases as powers of 2:** - $16 = 2^4$ - $\frac{1}{2} = 2^{-1}$ - $4 = 2^2$ So the equation becomes: $$\left(2^4\right)^x \times \left(2^{-1}\right)^x = \left(2^2\right)^{x+3}$$ 3. **Apply the power of a power rule:** $$2^{4x} \times 2^{-x} = 2^{2(x+3)}$$ 4. **Combine the left side using the product of powers rule:** $$2^{4x + (-x)} = 2^{2x + 6}$$ $$2^{3x} = 2^{2x + 6}$$ 5. **Since the bases are equal, set the exponents equal:** $$3x = 2x + 6$$ 6. **Solve for $x$:** $$3x - 2x = 6$$ $$\cancel{3x} - \cancel{2x} = 6$$ $$x = 6$$ **Final answer:** $$x = 6$$