<s>[INST] 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: $[0, \infty]$ ∪ ${-2}$
 [/INST] If $|x+1|\ge 0$ the equation becomes
$$
2\cdot 2^x-|2^x-1|=2^x+1
$$
that is equivalent to
$$
\begin{cases}
2^x-1\ge0\\
2\cdot 2^x-2^x+1=2^x+1
\end{cases}
\quad \lor\quad
\begin{cases}
2^x-1<0\\
2\cdot 2^x+2^x-1=2^x+1
\end{cases}
$$
can you solve these two systems? and solve also the other case: for $|x+1|< 0$ ?
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