Question

Genteeeee, me ajudem aqui pfvv

Answer 1

Resposta:

Explicação passo a passo:

$\Big(\frac{\sqrt{3}}{9}\Big)^{2x-2} =\frac{1}{27}\\\Big(\frac{3^{\frac{1}{2}}}{3^2}\Big)^{2x-2} =\frac{1}{3^3}\\\\\Big(3^{\frac{1}{2}-2}\Big)^{2x-2} =\Big(\frac{1}{3}\Big)^3\\\\\Big(3^{\frac{1}{2}-\frac{2}{2}}\Big)^{2x-2} =3^{-3}\\\\\Big(3^{\frac{1-2}{2}}\Big)^{2x-2} =3^{-3}\\\\\Big(3^{\frac{-1}{2}}\Big)^{2x-2} =3^{-3}\\\\3^{\frac{-1}{2}\cdot(2x-2)} =3^{-3}\\\\3^{-x+1} =3^{-3}$

$-x+1=-3\\-x=-3-1\\-x=-2\\\\x=\frac{-2}{-1}\\x=2$