Answer:
[tex] {8}^{x} \times {16}^{x } - 1 = 1 \\ {2}^{3x} \times {2}^{4x} = 1 + 1 \\ {2}^{7x} = {2}^{1} \\ 7x = 1 \\ x = \frac{1}{7} [/tex]
