Question

Solve the following inequality:
${\log }_{1/x}(2.5x-1)⩾-2$

Answer 1

$\frac{\log (2.5x-1)}{-\log x}\ge -2\frac{\log (2.5x-1)}{\log x}\le 2$

$2.5x-1>0x>\frac{1}{2.5}x\in (0.4,\infty )$ and $x>0$

Taking intersection, Hence $x\in (0.4,\infty )$

For $x\in (0.4,1),\log x<0$

$\log (2.5x-1)\ge 2\log x(2.5x-1)\ge x^2$

$\Rightarrow x^2-2.5x+1\le 0x\in [0.5,2]$

Hence, $x\in [0.5,1)$

For $x\in (1,\infty )\log x>0$

$\log (2.5x-1)\le 2\log x2.5x-1\le x^2{x}^2-2.5x+1\ge 0x\in (-\infty ,0.5]\cup [2,\infty )$

Hence, $x\in [2,\infty )$

Taking union we get $x\in [0.5,1)\cup [2,\infty )$