Question

me ajudem com essa exponencial:

Answer 1

$8^{x-1}= \sqrt[3]{2}\\ (2^3)^{x-1}= \sqrt[3]{2^1}\\ \not2^{3x-3} =\not2^{ \tfrac{1}{2} } \\\\ 3x-3= \dfrac{1}{2}\\\\ 2\cdot(3x-3)=1\\ 6x-6=1\\ 6x=7\\\\ x= \dfrac{7}{6} \\\\\\ \text{S}=\left\{ \dfrac{7}{6}\right\}$