Question

решить неравенство
$(\frac{1}{8} ) ^{2x+1/3} \ \textgreater \ 4$

Answer 1

$\Big(\dfrac{1}{8}\Big)^{\dfrac{2x+1}{3} } >4 \\\\ \Big((2^{-3})\Big)^{\dfrac{2x+1}{3} }>2^{2}\\\\\Big2^{-(2x+1)} >\Big2^{2}\\\\-2x-1>2\\\\-2x>3\\\\x<-1,5\\\\Otvet:\boxed{x\in(-\infty \ ; \ -1,5)}$