Question

matemática me ajudem nessa

Answer 1

3) $4^{2-3x}=\sqrt{2}$

$(2^2)^{2-3x}=2^{\frac{1}{2}}$

$2^{4-6x}=2^{\frac{1}{2}}$

$4-6x=\dfrac{1}{2}$

$8-12x=1$

$12x=7$

$\boxed{x=\dfrac{7}{12}}$

$\dfrac{7}{12}\in[0,1]$, pois $0<\dfrac{7}{12}<1$

Letra B