Question

Solution set of the inequality ${\left({\log }_{2}x\right)}^4-{\left({\log }_{1/2}\frac{x^3}{8}\right)}^2+9{\log }_2\left(\frac{32}{x^2}\right)<4{\left({\log }_{1/2}x\right)}^2$ is (a, b) $\cup$ (c, d) then the correct statement is

• A. $a=2b$ and $d=2c$
• B. $b=2a$ and $d=2c$
• C. ${\log }_{c}d={\log }_{b}a$
• D. there are 4 integers in $(c,d)$

Answer 1

Answer: (B), (C)

${\left({\log }_{2}x\right)}^4-{\left({\log }_2{\left(\frac{x}{2}\right)}^3\right)}^2+9\left[{\log }_{2}32-{\log }_2{x}^2\right]<4{\left({\log }_{2}x\right)}^2($ note $x>0)$
${\left({\log }_{2}x\right)}^4-{\left(3{\log }_{2}x-3\right)}^2+45-18{\log }_{2}x<4{\left({\log }_{2}x\right)}^2$
$\text{ let }{\log }_{2}x=t$
$t^4-(3t-3{)}^2+45-18t<4t^2\Rightarrow t^4-\left(9t^2+9-18t\right)-18t+45<4t^2$
$\Rightarrow t^4-13t^2+36<0\Rightarrow \left(t^2-4\right)\left(t^2-9\right)<0$
$\Rightarrow 4<t^2<9$
$t^2<9\Rightarrow -3<t<3$
and $t^2>4\Rightarrow t>2$ or $t<-2$
hence, $t\in (-3,-2)\cup (2,3)$
$x\in \left(\frac{1}{8},\frac{1}{4}\right)\cup (4,8)\Rightarrow B,C$