How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $

Question

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 $ ?

Answer 1

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 $\{-2\}\cup [0,+\infty)$.

Answer 2

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.

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There are two more cases to do: $-1\leq x<0$ and $x<-1$.

Answer 3

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

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