Solve for x:

$\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x-1} = \dfrac{27}{8}$

Given,

$\Rightarrow \Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x-1} = \dfrac{27}{8} \\[1em] \Rightarrow \Big[\Big({\dfrac{2}{3}}\Big)^\dfrac{x - 1}{3}\Big] = \dfrac{3^3}{2^3} \\[1em] \Rightarrow \Big({\dfrac{2}{3}}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big({\dfrac{2}{3}}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{2}{3}\Big)^{-3}$

Equating the exponents,

$\Rightarrow \dfrac{x - 1}{3} = -3 \\[1em] \Rightarrow x - 1 = -3 \times 3 \\[1em] \Rightarrow x - 1 = -9 \\[1em] \Rightarrow x = -9 + 1 = -8.$

Hence, x = -8.