1. **State the problem:** Simplify the expression $$\frac{\left(\frac{1}{2}\right)^x \cdot 8^x}{4^x}$$. 2. **Recall the laws of exponents:** - When multiplying with the same exponent, multiply the bases: $$a^x \cdot b^x = (a \cdot b)^x$$. - When dividing with the same exponent, divide the bases: $$\frac{a^x}{b^x} = \left(\frac{a}{b}\right)^x$$. 3. **Rewrite the bases as powers of 2:** - $$\frac{1}{2} = 2^{-1}$$ - $$8 = 2^3$$ - $$4 = 2^2$$ 4. **Substitute these into the expression:** $$\frac{(2^{-1})^x \cdot (2^3)^x}{(2^2)^x}$$ 5. **Apply the power of a power rule:** $$\frac{2^{-x} \cdot 2^{3x}}{2^{2x}}$$ 6. **Multiply the numerator terms:** $$2^{-x + 3x} = 2^{2x}$$ 7. **Divide numerator by denominator:** $$\frac{2^{2x}}{2^{2x}} = 2^{2x - 2x} = 2^0 = 1$$ **Final answer:** $$1$$