Question

Solve the following inequality:
${\log }_{x}2x⩽\sqrt{{\log }_x\left(2x^3\right)}$

Answer 1

${\left({\log }_x\left(2x\right)\right)}^2\le {\log }_x\left(2x^3\right){({\log }_{x}2+{\log }_{x}x)}^2\le {\log }_{x}2+3{\log }_{x}x{\log }_x^{2}2+1^2+2{\log }_{x}2\le {\log }_{x}2+3{\log }_{x}x{\log }_x^{2}2+1+2{\log }_{x}2\le {\log }_{x}2+3{\log }_x^{2}2+{\log }_{x}2-2\le 0{\log }_x^{2}2+2{\log }_{x}2-{\log }_{x}2-2\le 0{\log }_{x}2({\log }_{x}2+2)-1({\log }_{x}2+2)\le 0Hence{\log }_{x}2\in [-2,1]-2\le {\log }_{x}2\le 1-2{\left(\log x\right)}^2\le {\log }_{x}2\log x\le {\left(\log x\right)}^{2}2{\left(\log x\right)}^2+{\log }_{x}2\log x\ge 0\log x(2\log x+\log 2)\ge 0x\in (-\infty ,0]\cup [\frac{1}{4},\infty ){\left(\log x\right)}^2-{\log }_{x}2\log x\ge 0\log x(\log x-\log 2)\ge 0x\in (-\infty ,0]\cup [2,\infty )$

Taking intersection, we get $x\in [2,\infty )$