${2^\left|(x+2)\right|}$ - $\left|2^{x+1} -1 \right| $=$ 2^{x+1}+1$ .What is the Minimum value of ${x}$? I try to use logarithm but I  cannot be applied on the equation.
So how can I solve this  equation.
Any help will be appreciate.
 A: If $x \in [-1 , \infty)$,
then we have $|x+2|=x+2$ and $|2^{x+1}-1|=2^{x+1}-1$;
so in this interval, the equation has this form:
$$ 
\ \ \ \ \
2^{x+2}-(2^{x+1}-1)= 
2^{x+1}+1 
\Longleftrightarrow 
2.2^{x+1}-2^{x+1}+1= 
2^{x+1}+1 
\Longleftrightarrow 
\\ 
(2-1).2^{x+1}+1= 
2^{x+1}+1 
\Longleftrightarrow 
2^{x+1}+1= 
2^{x+1}+1 
; 
\ \ \ \ \
\checkmark 
\checkmark 
\checkmark 
$$
so we can conclude that 
every $x \in [-1 , \infty)$ satisfies the equation.  

If $x \in (\infty , -1]$, then we have 
$%% |x+2|=-(x+2)$ 
$|2^{x+1}-1|=2^{x+1}-1$;
so in this interval, the equation has this form:
$$ 
2^{|x+2|}-\Big(-(2^{x+1}-1)\Big)= 
2^{x+1}+1 
\Longleftrightarrow 
2^{|x+2|}=2 
\Longleftrightarrow 
|x+2|=1; 
$$ 
which gives us the new solution $x=-3 \in [-1 , \infty)$.  

So by the above we can conclude that all solutions are the following: 
$$ 
\{ -3 \} 
\cup 
[-1 , \infty) 
. 
$$
A: It seems that the absolute value signs naturally break this up into several cases: $x < -2$ vs. $x \geq -2$ and $2^{x+1} -1 \geq 0$ vs. $2^{x+1} -1 < 0$.
I would recommend finding out exactly what intervals of $x$ you are concerned with to get rid of the absolute value bars.
Then I would note that $2^{x+c} = 2^x2^c$ for any $c$, so you can try to solve for $2^x$.
