Question

$8^{2X+1} =\sqrt[3]{4^{X-1} }$

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle\\8^{2x+1} =\sqrt[3]{4^{x-1}}\\2^{3.(2x+1)} =4^{\frac{x-1}{3}}\\2^{3.(2x+1)} =2^{2(\frac{x-1}{3})}\\\\3(2x+1)=\frac{2(x-1)}{3}\\9(2x+1)=2x-2\\18x+9=2x-2\\16x=-11\\x=-\frac{11}{16}$