Question

Solve the following inequality:
${\log }_{2x+3}{x}^2<1$

Answer 1

$\frac{\log x^2}{\log 2x+3}<1$ and $x>0$

For $x>-1$

$\log x^2<\log (2x+3)x^2<2x+3x^2-2x-3<0(x-3)(x+1)<0x\in (-1,3)$

For $x\in (-3/2,-1)$

$\log x^2>\log (2x+3)x^2>2x+3x^2-2x-3>0(x-3)(x+1)>0x\in (-\infty ,-1)\cup (3,\infty )$

Hence $x\in (-3/2,-1)$

Taking intersection, we get $x\in (0,3)$