(5) \( \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n}}{16^{n-1} \cdot 3^{-n}} \)

To simplify the expression: \[ \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n}}{16^{n-1} \cdot 3^{-n}} \] **Step 1: Express All Numbers in Terms of Prime Factors** - \(8 = 2^3\) - \(6 = 2 \cdot 3\) - \(9 = 3^2\) - \(16 = 2^4\) **Step 2: Rewrite Each Term Using Prime Factors** \[ \begin{align*} 8^n &= (2^3)^n = 2^{3n} \\ 6^{n-3} &= (2 \cdot 3)^{n-3} = 2^{n-3} \cdot 3^{n-3} \\ 9^{1-n} &= (3^2)^{1-n} = 3^{2(1-n)} = 3^{2 - 2n} \\ 16^{n-1} &= (2^4)^{n-1} = 2^{4(n-1)} = 2^{4n-4} \\ 3^{-n} &= 3^{-n} \end{align*} \] **Step 3: Substitute Back into the Original Expression** \[ \frac{2^{3n} \cdot 2^{n-3} \cdot 3^{n-3} \cdot 3^{2-2n}}{2^{4n-4} \cdot 3^{-n}} \] **Step 4: Combine Like Terms** - For the base \(2\): \[ 2^{3n + n - 3} = 2^{4n - 3} \] - For the base \(3\): \[ 3^{n-3 + 2 - 2n} = 3^{-n - 1} \] So, the expression becomes: \[ \frac{2^{4n-3} \cdot 3^{-n-1}}{2^{4n-4} \cdot 3^{-n}} \] **Step 5: Simplify by Subtracting Exponents** - For the base \(2\): \[ 2^{4n-3} \div 2^{4n-4} = 2^{(4n-3) - (4n-4)} = 2^{1} = 2 \] - For the base \(3\): \[ 3^{-n-1} \div 3^{-n} = 3^{(-n-1) - (-n)} = 3^{-1} = \frac{1}{3} \] **Step 6: Combine the Results** \[ 2 \times \frac{1}{3} = \frac{2}{3} \] **Final Answer:** \[ \frac{2}{3} \]