Are you good in basic log's

We first let $\log _{ 2 }{ x } =y$ . Then the given inequality becomes: $\sqrt { y-1 } -\frac { 3 }{ 2 } y+2>0$ or $\sqrt { y-1 } >\frac { 3 }{ 2 } y-2$ Now here is an important thing to note..that we can square both sides of an inequality keeping its sign unchanged only if both sides are non-negative. Observing from the LHS of the above inequality we see that $y\ge 1$ . Also see that in that case..since LHS is under radical sign so it must be talking of the principle square root which is non-negative. Further we can say that the RHS is non-negative if $y\ge \frac { 4 }{ 3 }$ . So for $y\ge \frac { 4 }{ 3 }$ we can say both sides of the inequality are non negative and we can square it without reversing its sign. Squaring we get: $y-1>{ \left( \frac { 3 }{ 2 } y-2 \right) }^{ 2 }\\ \Rightarrow 9{ y }^{ 2 }-28y+20<0\\ \Rightarrow \left( y-2 \right) \left( 9y-10 \right) <0\\ \Rightarrow \frac { 10 }{ 9 } <y<2\\ \Rightarrow 2<\log _{ 2 }{ \left( \frac { 10 }{ 9 } \right) } <x<4$ There exists an integral solution $x=3$ in this region. But the question is whether there exists any integral solution outside this region (we can't say yet, because in order to square the inequality without reversing its sign we have added an extra condition to make its both sides non negative). To check that we have to look for solutions in the region $[1,\frac { 4 }{ 3 })$ , since the LHS becomes undefined for $y<1$ . We see that $\log _{ 2 }{ 2 } =1<\frac { 4 }{ 3 }$ So, we are left with only one option $x=2$ . We check that $x=2$ satisfies the above inequality. So the possible integer solutions are $x=2$ and $x=3$ and their sum is $5$ .