\( \frac{\sqrt{8^{x-2}} \cdot \sqrt[3]{9^{x}}}{\sqrt[3]{6^{2 x-9}} \cdot 2^{-\frac{x}{6}}} \)

To simplify the expression \( \frac{\sqrt{8^{x-2}} \cdot \sqrt[3]{9^{x}}}{\sqrt[3]{6^{2 x-9}} \cdot 2^{-\frac{x}{6}}} \), let's break it down step-by-step. 1. **Rewrite the terms with exponents:** - For \( \sqrt{8^{x-2}} \): \[ \sqrt{8^{x-2}} = (8^{x-2})^{1/2} = 8^{\frac{x-2}{2}} = (2^3)^{\frac{x-2}{2}} = 2^{\frac{3(x-2)}{2}} = 2^{\frac{3x - 6}{2}}. \] - For \( \sqrt[3]{9^{x}} \): \[ \sqrt[3]{9^{x}} = (9^{x})^{1/3} = 9^{\frac{x}{3}} = (3^2)^{\frac{x}{3}} = 3^{\frac{2x}{3}}. \] - For \( \sqrt[3]{6^{2x-9}} \): \[ \sqrt[3]{6^{2x-9}} = (6^{2x-9})^{1/3} = 6^{\frac{2x-9}{3}} = (2 \cdot 3)^{\frac{2x-9}{3}} = 2^{\frac{2x-9}{3}} \cdot 3^{\frac{2x-9}{3}}. \] - For \( 2^{-\frac{x}{6}} \): \[ 2^{-\frac{x}{6}} = \frac{1}{2^{\frac{x}{6}}}. \] 2. **Combine the terms:** The expression becomes: \[ \frac{2^{\frac{3x - 6}{2}} \cdot 3^{\frac{2x}{3}}}{2^{\frac{2x-9}{3}} \cdot 3^{\frac{2x-9}{3}} \cdot 2^{-\frac{x}{6}}}. \] 3. **For the base \( 2 \):** Let's simplify the powers of 2: \[ \text{Numerator: } 2^{\frac{3x - 6}{2}}. \] \[ \text{Denominator: } 2^{\frac{2x-9}{3}} \cdot 2^{-\frac{x}{6}} = 2^{\frac{2x-9}{3} - \frac{x}{6}}. \] Finding a common denominator (which is 6): \[ \frac{2x - 9}{3} = \frac{4x - 18}{6} \quad \text{and} \quad -\frac{x}{6}. \] Thus, \[ \frac{4x - 18 - x}{6} = \frac{3x - 18}{6}. \] Now the powers for \( 2 \): \[ 2^{\frac{3x - 6}{2} - \frac{3x - 18}{6}}. \] Compute the two powers: The LCM of the denominators is 6: \[ \frac{3x - 6}{2} = \frac{9x - 18}{6}, \quad \text{and thus: } \frac{9x - 18 - (3x - 18)}{6} = \frac{6x}{6} = x. \] Therefore, we have: \[ 2^{x}. \] 4. **For the base \( 3 \):** The numerator gives \( 3^{\frac{2x}{3}} \) and the denominator gives: \[ 3^{\frac{2x - 9}{3}} \implies \text{Power of } 3 = \frac{2x}{3} - \frac{2x - 9}{3} = \frac{2x - (2x - 9)}{3} = \frac{9}{3} = 3. \] 5. **Final Expression:** Now we can combine the results: \[ 2