Question

The set of real values of $x$ for which
${\log }_{(x+3)}(x^2-x-11)<0\text{if}x>-2$ is

Answer 1

Since $x>-2$
$\Rightarrow x+3\text{will always be greater than}1$

So, base of the logarithm is greater than $1$,
then inequality is equivalent to
We heve given
${\log }_{(x+3)}(x^2-x-11)<0$

${\log }_{(x+3)}(x^2-x-11)<{\log }_{(x+3)}1$
So,

$\Rightarrow x^2-x-11<1$

$\Rightarrow x^2-x-12<0$

$\Rightarrow (x-4)(x+3)<0$

$\Rightarrow x\in (-3,4)$

Given, $x>-2$

$\therefore x\in (-2,4)$

Hence the correct answer is Option C.