How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
 A: Consider three cases:.
i) $\bf x\leq -1$. Then $x\leq -1$, $2^x< 1$, and the equation becomes
$\frac{1}{2}2^{-x} - 2^x = 1- 2^x + 1$, that is $2^{-x}=4$ which implies that $x=-2\leq -1$.
ii) $\bf -1<x<0$. Then $x>-1$, $2^x< 1$, and the equation becomes $2^x=2\cdot 2^{x} - 2^x = 1- 2^x + 1=2-2^x$, that is $2^x=1$ which is impossible for $-1<x<0$.
iii) $\bf 0\leq x$. Then $x>-1$, $2^x\geq 1$, and the equation becomes $2^x=2^x$ which holds for all $x\geq 0$.
Hence the complete set of solutions is $\{-2\}\cup [0,+\infty)$.
A: If $x\geq 0$, then


*

*$|x+1| = x+1$

*$|2^x-1| = 2^x-1$


So the equation becomes
$$2^{x+1}-2^x=2^x-1+1\\
2^{x+1} - 2\cdot 2^x=0\\
2^{x+1}-2^{x+1}=0\\
0=0$$
so it is always true.

There are two more cases to do: $-1\leq x<0$ and $x<-1$.
A: Hint :divide into cases.
$$\forall x\geq 0 \to R.H.S=L.H.S \\2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \to \\2^{x+1}-2^x=2^x-1+1\\2(2^x)-2^x=2^x \checkmark\\ x\in[0,\infty)$$ 
Then $$for \space -\leq x\leq0 \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\
\to 2^{x + 1 } - 2^x =  1-2^x + 1 \implies x=-2\\2^x=2-2^x \to x=0 $$
$$for \space  x\leq-1  \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\2^{-x-1}-2^x=1-2^x+1\\2^{-x-1}=2 \\\to -x-1=1 \to \\x=-2$$
