Question

Sum of all integral values of $x$ satisfying the inequality
$3^{\frac{5}{2}{\log }_3(12-3x)}-3^{{\log }_{2}x}>32$ is

Answer 1

$\left(3^{2.5{\log }_3(12-3x)}\right)-\left(3^{{\log }_{2}x}\right)>32$ $\Rightarrow 12-3x>0\&x>0$
$\Rightarrow x\in (0,4)$
$\therefore x\in \{1,2,3\}$

For $x=1$, we get the expression as:
$\left(3^{2.5{\log }_{3}9}\right)-\left(3^{{\log }_{2}1}\right)=3^5-3^0>32$

For $x=2$,
$\left(3^{2.5{\log }_{3}6}\right)-\left(3^{{\log }_{2}2}\right)=(6^5{)}^{1/2}-3^1>32$
We know that,
$6^2-3>32\therefore (6^5{)}^{1/2}-3^1>32$

For $x=3$,
$\left(3^{2.5{\log }_{3}3}\right)-\left(3^{{\log }_{2}3}\right)=3^{2.5}-3^{{\log }_{2}3}<32$
We know that,
$3^3=27<32\therefore 3^{2.5}-3^{{\log }_{2}3}<32$

Hence, the required values are $1,2$ and the required sum $=3$