Question

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

Answer 1

${\log }_{2x+3}{x}^2<1$

$\Rightarrow \frac{\log x^2}{\log 2x+3}<1$

$\Rightarrow \log x^2<\log 2x+3$

$\Rightarrow x^2<2x+3$

$\Rightarrow x^2-2x-3<0$

$\Rightarrow x^2-3x+x-3<0$

$\Rightarrow (x-3)(x+1)<0$

$(x-3)(x+1)<0$

$(x-3)(x+1)$ will be less than zero if and only if:-

Case I : $x-3>0$ and $x+1<0$

$\Rightarrow x>3$ and $x<-1$

Not possible

Case II : $x-3<0$ and $x+1>0$

$\Rightarrow x<3$ and $x>-1$

$\Rightarrow -1<x<3$

Solution: $-1<x<3$

Interval Notation $(-1,3)$.