Question

Solve the following inequality:
${\log }_3\frac{1+2x}{1+x}<1$

Answer 1

$\frac{1+2x}{1+x}<3$

$\frac{1+2x-3-3x}{1+x}<0$

$\frac{-2-x}{1+x}<0$

$\frac{x+2}{1+x}>0$

$either$

$x+2>0andx+1>0$

$x>-2andx>-1$

$therfore,x>-1$

$or$

$x+2<0andx+1<0$

$x<-2andx<-1$

$therefore,x<-2$

$therefore,$

$x\epsilon (-\infty ,-2)\bigcup (-1,\infty )$