\(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\) \(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0\) \(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0\) \(3^{(3x + 1 - x - 4x)} = 3^0\) \(3^{(1 - 2x)} = 3^0\) since the bases are equal, 1 - 2x = 0 - 2x = -1 x = \(\frac{1}{2}\)