Question

$\sqrt[3]{x + x^{2} }$Sabendo que log1/2 x = -3, calcule o valor de

Answer 1

$log(\frac{1}{2x})=-3 \\ \frac{1}{2x}={10}^{-3}$

$\frac{1}{2x}=\frac{1}{1000} \\ 2x=1000$

$x=\frac{1000}{2} \\ x=500$

$=\sqrt[3]{500+{500}^{2}}$

$\sqrt[3]{500(1+500)}=\sqrt[3]{{2}^{2}.{5}^{3}.501}$

$5\sqrt[3]{4.501}=5\sqrt[3]{2004}$