Question

me ajudem a resolver essa equação
$\sqrt[x + 4]{ {2}^{3x - 8 } } = {2}^{x - 5}$

Answer 1

$\sqrt[x+4]{2^{3x-8}} =2^{x-5}\\\\ 2^{\frac{3x-8}{x+4}}=2^{x-5}\\\\ \frac{3x-8}{x+4}=x-5\\\\ 3x-8=(x-5)(x+4)\\ 3x-8=x^2+4x-5x-20\\ x^2-4x-12=0\\ \Delta=(-4)^2-4.1.(-12)=16+48=64\\\\ x=\frac{-(-4)\pm\sqrt{64}}{2.1}\\\\ x=\frac{4\pm8}{2}\\\\ x_1=\frac{4+8}{2}=\frac{12}{2}=6\\\\ x_2=\frac{4-8}{2}=\frac{-4}{2}=-2$