1. **State the problem:** Express the expression $$\frac{2^n \times 4^{n-1}}{8^{n-2}}$$ as a power with a single base. 2. **Recall the bases and their relationships:** - Note that $4 = 2^2$ and $8 = 2^3$. 3. **Rewrite all terms with base 2:** - $4^{n-1} = (2^2)^{n-1} = 2^{2(n-1)} = 2^{2n-2}$ - $8^{n-2} = (2^3)^{n-2} = 2^{3(n-2)} = 2^{3n-6}$ 4. **Substitute back into the expression:** $$\frac{2^n \times 2^{2n-2}}{2^{3n-6}}$$ 5. **Use exponent rules:** - Multiply powers with the same base by adding exponents: $$2^n \times 2^{2n-2} = 2^{n + 2n - 2} = 2^{3n - 2}$$ - Divide powers with the same base by subtracting exponents: $$\frac{2^{3n - 2}}{2^{3n - 6}} = 2^{(3n - 2) - (3n - 6)} = 2^{3n - 2 - 3n + 6} = 2^{4}$$ 6. **Final answer:** $$\boxed{2^4}$$ This simplifies to $16$ but the problem asks for a power with a single base, so $2^4$ is the final expression.