(3) \( \frac{3^{x+3} \cdot 18^{x}}{6^{x+1} \cdot 9^{x+1}} \)

To simplify the expression \( \frac{3^{x+3} \cdot 18^{x}}{6^{x+1} \cdot 9^{x+1}} \), we first express all terms in the numerator and denominator in terms of powers of 3. 1. **Rewrite the bases**: - \( 18 = 2 \cdot 3^2 \) - \( 6 = 2 \cdot 3 \) - \( 9 = 3^2 \) 2. **Substituting**: - \( 18^x = (2 \cdot 3^2)^x = 2^x \cdot 3^{2x} \) - \( 6^{x+1} = (2 \cdot 3)^{x+1} = 2^{x+1} \cdot 3^{x+1} \) - \( 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x + 2} \) 3. **Rewrite the expression**: The expression now looks like this: \[ \frac{3^{x+3} \cdot (2^{x} \cdot 3^{2x})}{(2^{x+1} \cdot 3^{x+1}) \cdot 3^{2x+2}} \] 4. **Combine the powers in the numerator**: - In the numerator: \( 3^{x+3 + 2x} = 3^{3x + 3} \) - And the \( 2^x \) stays as is: \( 2^x \) 5. **Combine the denominator**: - In the denominator: \( 2^{x+1} \cdot 3^{x + 2x + 2} = 2^{x+1} \cdot 3^{3x + 2} \) 6. **Put it all together**: \[ = \frac{2^x \cdot 3^{3x + 3}}{2^{x + 1} \cdot 3^{3x + 2}} \] 7. **Simplify further**: - The \( 3^{3x + 3} \) over \( 3^{3x + 2} \) gives \( 3^1 = 3 \). - The \( \frac{2^x}{2^{x+1}} \) leads to \( \frac{2^x}{2^x \cdot 2} = \frac{1}{2} \). 8. **Final simplification**: \[ \frac{3}{2} \] Thus, the simplified expression is: \[ \frac{3}{2} \]