Solve the equation: $\;$ $\dfrac{2^{x - 1} \times 4^{1 - x}}{8^{x - 1}} = 64$
Given equation: $\;\;$ $\dfrac{2^{x - 1} \times 4^{1 - x}}{8^{x - 1}} = 64$ $\;\;\; \cdots \; (1)$
i.e. $\;$ $\dfrac{2^{x - 1} \times \left(2^2\right)^{1 - x}}{\left(2^3\right)^{x - 1}} = 2^6$
i.e. $\;$ $2^{\left(x - 1 + 2 - 2x - 3x + 3\right)} = 2^6$
i.e. $\;$ $2^{-4x + 4} = 2^6$
i.e. $\;$ $-4x + 4 = 6$
i.e. $\;$ $-4x = 2$
i.e. $\;$ $x = \dfrac{-1}{2}$