Question 4
*4. ${\left(\frac{\text{1}}{\text{3}}\right)}^{\frac{\text{|x+2|}}{\text{2-|x|}}}\text{>9}$

Dear Student,

$$\begin{gathered} {\left(\frac{1}{3}\right)}^{\frac{\left|x+2\right|}{2-\left|x\right|}}>9 \\ \Rightarrow {\left(3^{-1}\right)}^{\frac{\left|x+2\right|}{2-\left|x\right|}}>3^2 \\ \Rightarrow 3^{-\frac{\left|x+2\right|}{2-\left|x\right|}}>3^2 \\ \Rightarrow 3^{\frac{\left|x+2\right|}{\left|x\right|-2}}>3^2 \\ \Rightarrow \frac{\left|x+2\right|}{\left|x\right|-2}>2 \\ \Rightarrow \left|x+2\right|>2\left(\left|x\right|-2\right) \\ \Rightarrow \left|x+2\right|>2\left|x\right|-4 \\ \Rightarrow 4>2\left|x\right|-\left|x+2\right| \\ i.e. 2\left|x\right|-\left|x+2\right|<4 \\ For x<-2, \\ 2\left|x\right|-\left|x+2\right|=2\left(-x\right)-\left[-\left(x+2\right)\right]=-2x+x+2=-x+2>2+2=4 \\ i.e. 2\left|x\right|-\left|x+2\right|>4 \\ \therefore reject this case. \\ For x=-2 \\ 2\left|x\right|-\left|x+2\right|=2\left(2\right)-0=4 \\ i.e. 2\left|x\right|-\left|x+2\right|=4 \\ \therefore reject this case. \\ For -2<x<0 \\ 2\left|x\right|-\left|x+2\right|=2\left(-x\right)-\left(x+2\right)=-2x-x-2=-3x-2 \\ -2<x<0\Rightarrow \left(-3\right)\left(-2\right)>-3x>\left(-3\right)\left(0\right)\Rightarrow 6>-3x>0 \\ \Rightarrow 6-2>-3x-2>0-2 \\ \Rightarrow 4>2\left|x\right|-\left|x+2\right|>-2 \\ \therefore accept this case. \\ For x=0, \\ 2\left|x\right|-\left|x+2\right|=2\left(0\right)-2=-2<4 \\ \therefore accept this case. \\ For 0<x<6 \\ 2\left|x\right|-\left|x+2\right|=2x-\left(x+2\right)=2x-x-2=x-2 \\ 0<x<6\Rightarrow 0-2<x-2<6-2\Rightarrow -2<2\left|x\right|-\left|x+2\right|<4 \\ \Rightarrow 4>2\left|x\right|-\left|x+2\right|>-2 \\ \therefore accept this case. \\ For x\ge 6, \\ 2\left|x\right|-\left|x+2\right|=2x-\left(x+2\right)=2x-x-2=x-2 \\ x\ge 6\Rightarrow x-2\ge 6-2=4 \\ 2\left|x\right|-\left|x+2\right|\ge 4 \\ \therefore reject this case. \\ Thus solution set is \left(-2,6\right). \end{gathered}$$

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