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 $ ?

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 $$for \space -\leq x\leq0 \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\
\to 2^{x + 1 } - 2^x =  1-2^x + 1 \implies x=-2\\2^x=2-2^x \to x=0 $$
$$for \space  x\leq-1  \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\2^{-x-1}-2^x=1-2^x+1\\2^{-x-1}=2 \\\to -x-1=1 \to \\x=-2$$
