Question

Solve ${\log }_{(x+3)}(x^2-x)<1$, then $xϵ(a,b)$
What is $ab$?

Answer 1

${\log }_{x+3}(x^2-x)<1$
$x(x-1)>0\Rightarrow x>1$ or $x<0$
If $x+3>1\Rightarrow x>-2$
then $x^2-x<x+3$
or $x^2-2x-3<0$
or $(x-3)(x+1)<0$
Hence, $xϵ(-1,0)\cup (1,3)$
If $0<x+3<1-3<x<-2$, then
$x^2-x>x+3$
or $x^2-2x-3>0$
or $(x-3)(x+1)$
$\Rightarrow$ $xϵ(-3,-2)$
Ans: 6