How do you solve the equation $2^{|x+1|}-|2^x-1| = 2^x+1$? | ... | is absolute value
$2^{|x+1|}-|2^x-1| = 2^x+1$, move $|2^x-1|$ to RHS: $2^{|x+1|} = 2^x+1+|2^x-1|$
Log base 2 each side:
$|x+1| = log_2 (2^x+1+|2^x-1|)$
Here we quickly observe: $-1$ is not a solution but $0$ is a solution. If you try negative numbers $-2$ is also a solution.
How do I find the value of $x$ (not intuitively like I did previously) for which the equation is true?
Thanks in advance for any help.
edit: Solved, thanks to all who replied. Split $2^{|x+1|} = 2^x+1+|2^x-1|$ into three cases: $x >= 0$ and $x < -1$ and $-1 <= x < 0$. Solution: $[0, \infty]$ ∪ ${-2}$

For $x\geq0$ the equation becomes
$$2^{x+1}-(2^x-1)=2^x+1$$
and the left hand side simplifies to $2^x+1$ so that every $x\geq 0$ is a solution.
For $x<0$ the equation becomes
$$2^{|x+1|}+(2^x-1)=2^x+1$$
so we get $2^{|x+1|}=2$ so that $|x+1|=1$ so that $x=-2$ is another solution.
