Question

The set of real values of $x$ for which inequality
${\log }_2(x^2-x-6)+{\log }_{0.5}(x-3)<2{\log }_{2}3$
holds true is

Answer 1

${\log }_2(x^2-x-6)+{\log }_{0.5}(x-3)<2{\log }_{2}3$

Condtion 1: For expression to be defined
$x^2-x-6>0\text{and}x-3>0$

$\Rightarrow x^2-x-6>0\text{and}x-3>0$

$\Rightarrow x^2-3x+2x-6>0\text{and}x-3>0$

$\Rightarrow (x-3)(x+2)>0\text{and}x-3>0$
Common port between these two conditions $x\in (3,\infty )$

So, above given logarithmic expression is defined only in the interval of $x\in (3,\infty )\cdots \left(1\right)$

Condition 2:

Make same base for all logarithmic terms

${\log }_2(x^2-x-6)+{\log }_{0.5}(x-3)<2{\log }_{2}3$

$\Rightarrow {\log }_2(x-3)+{\log }_2(x+2)+{\log }_{0.5}(x-3)<2{\log }_{2}3$
$\Rightarrow {\log }_2(x-3)+{\log }_2(x+2)+{\log }_{2^{-1}}(x-3)<2{\log }_{2}3$
$\Rightarrow {\log }_2(x-3)+{\log }_2(x+2)-{\log }_2(x-3)<2{\log }_{2}3$
$\Rightarrow {\log }_2(x+2)<{\log }_{2}9$

Base of the log is greater than $1$,
then inequality is equivalent to
$x+2<9$
$\Rightarrow x<7$
$\Rightarrow x\in (-\infty ,7)\cdots \left(2\right)$

But from condition (1) of log we got, $x\in (3,\infty )$

Common port of equation $\left(1\right)$ and $\left(2\right)$ can be
calculated based on number line
$\therefore x\in (3,7)$

Hence the correct answer is Option B.