Question

The inequality $\sqrt{x^{{\log }_2\sqrt{x}}}\ge 2$ is satisfied by

• A. only one value of $x$
• B. $xϵ[0,\frac{1}{4}]$
• C. $xϵ$$(0,\frac{1}{4}]\bigcup$ [4, $\infty$)
• D. $1<x<2$

Answer 1

The correct option is C $xϵ$$(0,\frac{1}{4}]\bigcup$ [4, $\infty$)

$x^{{\log }_2\sqrt{x}}\ge 4$ (squaring both sides)

${\log }_2\sqrt{x}\ge {\log }_x\left(4\right){\log }_2\left(x\right)\ge 2{\log }_x\left(4\right){\log }_2\left(x\right)\ge 4\frac{\log 2}{\log x}{\left(\log x\right)}^2\ge 4{\left(\log 2\right)}^2{\left(\log x\right)}^2-4{\left(\log 2\right)}^2\ge 0[\log x+2\log 2][\log x-2\log 2]\ge 0[\log x+\log 4][\log x-\log 4]\ge 0\log x\in (-\infty ,-\log 4]\bigcup [\log 4,\infty )x\in (-0,\frac{1}{4}]\bigcup [4,\infty )$