\( \begin{array}{lll}\text { (a) } \frac{16^{2} \cdot\left(3^{2}\right)^{4}}{2^{7} \cdot 2^{5} \cdot 81} & \text { (b) } \quad \frac{\left(2^{x}\right)^{4} \cdot 9^{x}}{12^{2 x}} & \text { (c) } \frac{16^{x-1}}{2^{4 x} \cdot 32}\end{array} \)

Let's simplify each of the given expressions step by step. ### Problem (a) \[ \frac{16^{2} \cdot \left(3^{2}\right)^{4}}{2^{7} \cdot 2^{5} \cdot 81} \] **Step 1:** Express all terms with base 2 and base 3. - \(16 = 2^4\), so \(16^2 = (2^4)^2 = 2^8\) - \(3^2 = 9\), so \(\left(3^2\right)^4 = 3^{8}\) - \(81 = 3^4\) - \(2^{7} \cdot 2^{5} = 2^{12}\) **Step 2:** Substitute these back into the expression: \[ \frac{2^8 \cdot 3^8}{2^{12} \cdot 3^4} \] **Step 3:** Simplify by subtracting exponents: \[ 2^{8-12} \cdot 3^{8-4} = 2^{-4} \cdot 3^4 = \frac{3^4}{2^4} = \frac{81}{16} \] **Answer for (a):** \[ \frac{81}{16} \] --- ### Problem (b) \[ \frac{\left(2^{x}\right)^{4} \cdot 9^{x}}{12^{2x}} \] **Step 1:** Simplify each component. - \(\left(2^x\right)^4 = 2^{4x}\) - \(9 = 3^2\), so \(9^x = 3^{2x}\) - \(12 = 2^2 \cdot 3\), so \(12^{2x} = (2^2 \cdot 3)^{2x} = 2^{4x} \cdot 3^{2x}\) **Step 2:** Substitute these back into the expression: \[ \frac{2^{4x} \cdot 3^{2x}}{2^{4x} \cdot 3^{2x}} = 1 \] **Answer for (b):** \[ 1 \] --- ### Problem (c) \[ \frac{16^{x-1}}{2^{4x} \cdot 32} \] **Step 1:** Express all terms with base 2. - \(16 = 2^4\), so \(16^{x-1} = (2^4)^{x-1} = 2^{4x - 4}\) - \(32 = 2^5\) **Step 2:** Substitute these back into the expression: \[ \frac{2^{4x - 4}}{2^{4x} \cdot 2^5} = \frac{2^{4x - 4}}{2^{4x + 5}} = 2^{-9} = \frac{1}{512} \] **Answer for (c):** \[ \frac{1}{512} \] --- ### Summary of Simplified Expressions \[ \begin{align*} \text{(a)} & \quad \frac{81}{16} \\ \text{(b)} & \quad 1 \\ \text{(c)} & \quad \frac{1}{512} \end{align*} \]