Question

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

Answer 1

$(3^{\frac{5}{2}{\log }_3(12-3x)})-3^{{\log }_{2}x}>32\text{Now,}12-3x>0\Rightarrow x<4\text{Also,}x>0\Rightarrow 0<x<4\text{Possible values of}x=1,2,3\text{Checking for}x=1\text{LHS}=3^{\frac{5}{2}{\log }_{3}9}-3^{{\log }_{2}1}=3^5-1=242>32\text{Checking for}x=2\text{LHS}=3^{\frac{5}{2}{\log }_{3}6}-3^{{\log }_{2}2}=6^{\frac{5}{2}}-3>32\text{Checking for}x=3\text{LHS}=3^{\frac{5}{2}{\log }_{3}3}-3^{{\log }_{2}3}=(\sqrt{3}{)}^5-3^{3{\log }_{2}3}=9\sqrt{3}-3^{3{\log }_{2}3}<32\therefore x=1,2\Rightarrow \text{Sum}=3$