Solve equation $log_{\frac{1}{2}}\left | x \right |=\frac{1}{4}\left ( \left | x-2 \right |+\left | x+2 \right | \right )$ Solve equation $log_{\frac{1}{2}}\left | x \right |=\frac{1}{4}\left ( \left | x-2 \right |+\left | x+2 \right | \right )$
I tried solving 4 separate cases
1.) $x< -2$
2.) $x\in[ \,-2,0\rangle$
3.) $x\in[ \,0,2\rangle$
4.) $x\geqslant2$
But I dont know how to solve this
in first case I get
$2^{\frac{1}{2}x}=-x$
And I dont know how to solve this.

Since $x < -2$, we see that $2^{\frac12 x} < 2^0 = 1$ while $-x > 2$, so the equation is never satisfied. The case for $x \ge 2$ is very similar.
As to case 2, that is, $-2 \le x < 0$, we see that
$$\frac 14(|x-2|+|x+2|) = \frac14(2-x+x+2) = 1$$
and $\log_{\frac12}|x| = 1$ is easy to solve. Same goes for case 3: $0 \le x < 2$.
