Question

$2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1$
Find the value of x

Answer 1

Case 1:- Let,

$\small\longrightarrow |x+2|=x+2$

$\small\Longrightarrow x+2\in [0,\ \infty)$

$\small\longrightarrow x\in [-2,\ \infty)\quad\quad\dots(1)$

Let,

$\small\longrightarrow\left|2^{x+1}-1\right|=2^{x+1}-1$

$\small\Longrightarrow 2^{x+1}-1\in[0,\ \infty)$

$\small\longrightarrow 2^{x+1}\in[1,\ \infty)$

$\small\longrightarrow x+1\in[0,\ \infty)$

$\small\longrightarrow x\in[-1,\ \infty)\quad\quad\dots(i)$

Taking (1) and (i),

$\small\longrightarrow x\in[-1,\ \infty)$

Then,

$\small\longrightarrow 2^{x+2}-\left(2^{x+1}-1\right)=2^{x+1}+1$

$\small\longrightarrow 2^{x+2}-2^{x+1}+1=2^{x+1}+1$

$\small\longrightarrow 2^{x+2}-2^{x+1}-2^{x+1}=0$

$\small\longrightarrow 2^{x+2}-2\cdot2^{x+1}=0$

$\small\longrightarrow 2^{x+2}-2^{x+2}=0$

$\small\longrightarrow0=0$

So the equation is satisfied for every x in the domain.

$\small\Longrightarrow x\in[-1,\ \infty)\quad\quad\dots(a)$

Case 1.2:- Let,

$\small\longrightarrow\left|2^{x+1}-1\right|=1-2^{x+1}$

$\small\Longrightarrow x\in(-\infty,\ -1]\quad\quad\dots(ii)$

Taking (1) and (ii),

$\small\longrightarrow x\in[-2,\ -1]$

Then,

$\small\longrightarrow 2^{x+2}-\left(1-2^{x+1}\right)=2^{x+1}+1$

$\small\longrightarrow 2^{x+2}-1+2^{x+1}=2^{x+1}+1$

$\small\longrightarrow 2^{x+2}=2$

$\small\longrightarrow 2^{x+1}=1$

$\small\longrightarrow x+1=0$

$\small\longrightarrow x=-1\in[-2,\ -1]$

$\small\Longrightarrow x\in\{-1\}\quad\quad\dots(b)$

Case 2:- Let,

$\small\longrightarrow |x+2|=-x-2$

$\small\Longrightarrow x\in(-\infty,\ -2]\quad\quad\dots(2)$

Taking (2) and (i),

$\small\Longrightarrow x\in\phi\quad\quad\dots(c)$

Taking (2) and (ii),

$\small\Longrightarrow x\in(-\infty,\ -2]$

Then,

$\small\longrightarrow 2^{-x-2}-\left(1-2^{x+1}\right)=2^{x+1}+1$

$\small\longrightarrow 2^{-x-2}-1+2^{x+1}=2^{x+1}+1$

$\small\longrightarrow 2^{-x-2}=2$

$\small\longrightarrow -x-2=1$

$\small\longrightarrow x=-3\in(-\infty,\ -2]$

$\small\Longrightarrow x\in\{-3\}\quad\quad\dots(d)$

Taking $\small(a)\lor(b)\lor(c)\lor(d),$

$\small\longrightarrow x\in[-1,\ \infty)\cup\{-1\}\cup\phi\cup\{-3\}$

$\small\text{ \longrightarrow\underline{\underline{x\in\{-3\}\cup[-1,\ \infty)}} }$