\( \begin{array}{ll}\text { d) } \frac{81^{a+1} \cdot 9^{3-a}}{7^{a+2} \cdot 3^{1-a}} & \text { e) } \frac{32^{x} \cdot 2^{2-x}}{8^{x+1} \cdot 2^{x}}\end{array} \)

Let's simplify each expression step by step. --- ### **d)** \[ \frac{81^{a+1} \cdot 9^{3-a}}{7^{a+2} \cdot 3^{1-a}} \] **Step 1: Express all terms with base 3.** - \(81 = 3^4\) - \(9 = 3^2\) So, \[ 81^{a+1} = (3^4)^{a+1} = 3^{4(a+1)} = 3^{4a + 4} \] \[ 9^{3-a} = (3^2)^{3-a} = 3^{2(3-a)} = 3^{6 - 2a} \] **Step 2: Combine the numerator.** \[ 3^{4a + 4} \cdot 3^{6 - 2a} = 3^{(4a + 4) + (6 - 2a)} = 3^{2a + 10} \] **Step 3: Combine the denominator.** \[ 7^{a+2} \cdot 3^{1 - a} \] **Step 4: Simplify the entire expression.** \[ \frac{3^{2a + 10}}{7^{a+2} \cdot 3^{1 - a}} = \frac{3^{2a + 10}}{3^{1 - a}} \cdot \frac{1}{7^{a+2}} = 3^{(2a + 10) - (1 - a)} \cdot 7^{-(a+2)} = 3^{3a + 9} \cdot 7^{-(a+2)} \] \[ = \frac{3^{3a + 9}}{7^{a + 2}} \] --- ### **e)** \[ \frac{32^{x} \cdot 2^{2-x}}{8^{x+1} \cdot 2^{x}} \] **Step 1: Express all terms with base 2.** - \(32 = 2^5\) - \(8 = 2^3\) So, \[ 32^{x} = (2^5)^x = 2^{5x} \] \[ 8^{x+1} = (2^3)^{x+1} = 2^{3(x+1)} = 2^{3x + 3} \] **Step 2: Combine the numerator.** \[ 2^{5x} \cdot 2^{2 - x} = 2^{5x + 2 - x} = 2^{4x + 2} \] **Step 3: Combine the denominator.** \[ 2^{3x + 3} \cdot 2^{x} = 2^{3x + 3 + x} = 2^{4x + 3} \] **Step 4: Simplify the entire expression.** \[ \frac{2^{4x + 2}}{2^{4x + 3}} = 2^{(4x + 2) - (4x + 3)} = 2^{-1} = \frac{1}{2} \] --- ### **Final Simplified Forms:** - **d)** \(\displaystyle \frac{3^{3a + 9}}{7^{a + 2}}\) - **e)** \(\displaystyle \frac{1}{2}\)