Find the solution to the given equation ${{2}^{\left| x+2 \right|}}-\left| {{2}^{x+1}}-1 \right|={{2}^{x+1}}+1$. with certain conditions at which the equation will satisfy.

Hint:To find the absolute value of any equation, it has two values one is less than zero, and another one is greater than zero i.e.,$x=a\ And x=-a$.We need to make them two separate equations and then solve them separately. If we found any inequality in that case there will be no solutions.

Complete step by step answer:

We have given in the question that,
${{2}^{\left| x+2 \right|}}-\left| {{2}^{x+1}}-1 \right|={{2}^{x+1}}+1$
As it is given mod we should consider three cases here that is,
Case(1): For $x\ge 0$
The given equation will become as follows,
⟹${{2}^{\left| x+2 \right|}}-\left| {{2}^{x+1}}-1 \right|={{2}^{x+1}}+1$
⟹${{2}^{\left( (x+1)+1 \right)}}-({{2}^{x+1}}-1)={{2}^{x+1}}+1$
(Since for $x\ge 0$ mod value is always positive)
Now ,
⟹${{2}^{x+2}}-{{2}^{x+1}}+1={{2}^{x+1}}+1$
+1 and +1 on both sides will be cancelled out and we get
⟹${{2}^{x+2}}={{2}^{x+1}}+{{2}^{x+1}}$
⟹${{2}^{x+2}}={{2.2}^{x+1}}$
On simplifying we get as follows
⟹${{2}^{x+2}}={{2}^{1+x+1}}$
As we know if bases are equal then their powers also equal.
⟹$x+2=x+2$
⟹$2=2$ which is true
This means that the given equation has solutions for all $x\ge 0$
Case(2):For $-1\le x<0$the given equation will become as
⟹${{2}^{\left| x+2 \right|}}-\left| {{2}^{x+1}}-1 \right|={{2}^{x+1}}+1$
Here the mod becomes negetive as $-1\le x<0$
⟹${{2}^{x+2}}+{{2}^{x+1}}-1={{2}^{x+1}}+1$
Here, ${{2}^{x+1}}$ terms on both sides will get cancelled out
On simplifying the above equations we get as follows,
⟹${{2}^{x+2}}={{2}^{1}}$
As the bases are equal their exponents are also equal
⟹$x+2=1$
⟹$x=-1$
Hence $x=-1$ for $-1\le x<0$
Case(3):For $x<-1$ the given equations will become as follows
⟹${{2}^{\left| x+2 \right|}}-\left| {{2}^{x+1}}-1 \right|={{2}^{x+1}}+1$
⟹${{2}^{-x-2}}+{{2}^{x+1}}-1={{2}^{x+1}}+1$
On simplifying we get
⟹${{2}^{-x-2}}={{2}^{1}}$
As the bases “2” are equal their exponents are also equal
⟹$-x-2=1$
On simplifying this arthmetic expression we get
⟹$x=-3$
Hence for $x<-1$ the given equation will satisfy .


Note:
 Here the students will do mistakes in considering the mod as positive and negative in different conditions. And also the least value of the given equation will obtain when $x=-3$ and also we should be careful while solving the equation and equate the powers when bases are equal.