1. **Problem:** Simplify the expression $\sqrt{x^5}$. 2. **Formula and rules:** For square roots, $\sqrt{a^m} = a^{\frac{m}{2}}$. When the exponent inside the root is odd, separate the power into an even part and a leftover 1 power: $$\sqrt{x^5} = \sqrt{x^4 \cdot x} = \sqrt{x^4} \cdot \sqrt{x}$$ Since $\sqrt{x^4} = x^{\frac{4}{2}} = x^2$, and $\sqrt{x}$ remains under the root. 3. **Intermediate work:** $$\sqrt{x^5} = \sqrt{x^4} \cdot \sqrt{x} = x^2 \sqrt{x}$$ 4. **Explanation:** We break down the exponent 5 into 4 (which is even) and 1. The square root of $x^4$ is $x^2$ because $\sqrt{x^4} = x^{4/2} = x^2$. The leftover $x$ stays inside the root. 5. **Final answer:** $$\boxed{x^2 \sqrt{x}}$$