Question

8^2x+1=∛$4^x-1$

Answer 1

$8^{2x+1}= \sqrt[3]{4^{x-1} }\\ (2^3)^{2x+1}= \sqrt[3]{(2^{2})^{x-1} }\\ 2^{6x+3}= \sqrt[3]{2^{2x-2} }\\\\ 2^{6x+3}=2^{ \tfrac{2x-2}{3} }\\\\ \not2^{6x+3}=\not2^{ \tfrac{2x-2}{3} }\\\\ 6x+3= \dfrac{2x-2}{3}\\\\ 3(6x+3)=2x-2\\ 18x+9=2x-2\\ 18x-2x=-2-9\\ 16x=-11\\\\ x=- \dfrac{11}{16}\\\\ S=\left\{- \dfrac{11}{16}\right\}$

Espero ter ajudado e tenha ótimos estudos =))