(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 start by rewriting the bases in terms of powers of 3 and 2: 1. Rewrite \( 18 \), \( 6 \), and \( 9 \): - \( 18 = 2 \cdot 3^2 \) - \( 6 = 2 \cdot 3 \) - \( 9 = 3^2 \) Now substitute these into the expression: \[ 18^x = (2 \cdot 3^2)^x = 2^x \cdot 3^{2x} \] \[ 6^{x+1} = (2 \cdot 3)^{x+1} = 2^{x+1} \cdot 3^{x+1} \] \[ 9^{x+1} = (3^2)^{x+1} = 3^{2x+2} \] Now, substitute everything back into the original expression: \[ \frac{3^{x+3} \cdot (2^x \cdot 3^{2x})}{(2^{x+1} \cdot 3^{x+1}) \cdot 3^{2x+2}} \] Now simplify it step by step: The numerator becomes: \[ 3^{x+3} \cdot 2^x \cdot 3^{2x} = 2^x \cdot 3^{(x+3) + 2x} = 2^x \cdot 3^{3x + 3} \] The denominator becomes: \[ (2^{x+1} \cdot 3^{x+1}) \cdot 3^{2x+2} = 2^{x+1} \cdot 3^{(x+1) + 2x + 2} = 2^{x+1} \cdot 3^{3x + 3} \] Now, rewriting the whole expression, we have: \[ \frac{2^x \cdot 3^{3x + 3}}{2^{x+1} \cdot 3^{3x + 3}} \] Since \(3^{3x + 3}\) in both the numerator and the denominator cancels out: \[ \frac{2^x}{2^{x+1}} = \frac{1}{2} \] Thus, the simplified expression is: \[ \frac{1}{2} \]