Question

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

Answer 1

$\left(3^{\frac{5}{2}{\log }_3(12-3x)}\right)-\left(3^{{\log }_{3}x}\right)>32$ is valid only when

$12-3x>0\Rightarrow x<4$ ... $\left(i\right)$

Also, $x>0$ ... $\left(ii\right)$

Now, $\left(3^{\frac{5}{2}{\log }_3(12-3x)}\right)-3^{{\log }_{3}x}>32$

$\Rightarrow 3^{{\log }_3(12-3x{)}^{\frac{5}{2}}}-3^{{\log }_{3}x}>32$

$\Rightarrow \left\{\right(12-3x{)}^{\frac{5}{2}}\}-x>32\cdots \cdots \{\because a^{{\log }_{a}b}=b\}$

$\Rightarrow (12-3x{)}^{\frac{5}{2}}-32>x$

$\Rightarrow (12-3x{)}^{\frac{5}{2}}>x+32$

$\Rightarrow (12-3x{)}^5>(x+32{)}^2\cdots \cdots \left(iii\right)$

from $\left(i\right),\left(ii\right)$,

$xϵ(0,4)$
there are $3$ integral values in this range, we will check them with equation $\left(iii\right)$

So, we have to check for $x=1,2,3$

$\therefore x=1\Rightarrow [9^5>(33{)}^2\equiv 9\times {81}^2>{33}^2]\to$ True

$\therefore x=2\Rightarrow [6^5>(34{)}^2\equiv 6\times {36}^2>{34}^2]\to$ True

$\therefore x=3\Rightarrow [3^5>(35{)}^2\equiv 243>1125]\to$ False

$\therefore x=1,2$

Sum of integral values of $x$ satisfying given equation is $1+2=3$.