Question

Solve the following inequality:
$x^{{\log }^{2}x-3\log x+1}>1000$

Answer 1

${}_x{log}^{2}x-3logx+1>1000\Rightarrow ({log}^{2}x-3logx+1)logx>3\Rightarrow (logx{)}^3-3{\left(logx\right)}^2+logx-3>0\Rightarrow \left[\right(logx{)}^2+1\left]\right(logx-3)>0$

$1+{\left(logx\right)}^2$ is always positive.

Therefore, $logx-3>0x>1000x\in (1000,\infty ).$