Question

Solve the following inequality:
$({\log }_{1/2}x{)}^2-{\log }_{1/2}{x}^2>({\log }_{1/2}3{)}^2-1$

Answer 1

${\left({\log }_{1/2}x\right)}^2-{\log }_{1/2}{x}^2>{\left({\log }_{1/2}3\right)}^2-1$

${\left({\log }_{2}x\right)}^2+2{\log }_{2}x-{\left({\log }_{2}3\right)}^2+1>0$

${\log }_{2}xϵ(-\infty ,-1-{\log }_{2}3)\bigcup (-1+{\log }_{2}3,\infty )$

$xϵ(0,2^{-1}{2}^{-{\log }_{2}3})\bigcup (2^{-1}{2}^{{\log }_{2}3},\infty )$
$xϵ(0,\frac{1}{6})\bigcup (\frac{3}{2},\infty )$