Solve equation $log_{\frac{1}{2}}\left | x \right |=\frac{1}{4}\left ( \left | x-2 \right |+\left | x+2 \right | \right )$

Question

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.

In general, any equation in the form $a^x = bx$ cannot be solved in a standard way. — the_candyman, Oct 21 '20 at 15:46

score 2 · Accepted Answer · answered Oct 21 '20 at 15:50

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: $0 \le x < 2$.

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