How to prove this inequality with some condition?

$ x $ must be strictly positive, so, whenever $ x > e $, by simplifying, the given equation can be in the form of $ln(x) > 2 $, which is true for $ x > e^2 $.
And for $0 < x < e $, then the given equation becomes $ln(x) < 2 $, which is true for $ 0 < x < e^2 $.
So, ultimately, the given inequality is possible for $0 < x < e $ and $ x > e^2 $.( Since , $0 < x < e $ is the common region between $0 < x < e $ and $0 < x < e^2 $, satisfying the given inequality , on the other hand, $ x > e^2 $ is common region between $ x > e^2 $ and $ x > e $, satisfying the given inequality.)