Transforming absolute-value sign to plus-minus sign

Question

I recently encountered this algebra transformation from an absolute-value sign to plus-minus sign:

$$|y|=5e^{2x-2}\iff y=\pm5e^{2x-2}$$

I am unable to get my head around how it really works. What is the underlying principle that justifies such transformation?

Substitute in the $y = ...$ equation into the original one with the absolute values. Try both the plus and minus versions, and see what happens. — PrincessEev, Dec 04 '18 at 02:53

score 4 · Accepted Answer · answered Dec 04 '18 at 02:52

4

If $|x|=1$, then it is either $x=1$ or $x=-1$.

Hence if $|x|=y$, then it is possible that $x=y$ or $x=-y$.

answered Dec 04 '18 at 02:52

Siong Thye Goh

score 0 · Answer 2 · answered Mar 05 '23 at 18:18

Adding to the accepted answer:

$$|y|=5e^{2x-2}\iff y=\pm5e^{2x-2}$$

In general, $$|y|=f\quad\iff\quad y=\pm f\:\:\text{and}\:\:\color{brown}{f\ge0}.$$

The condition $\color{brown}{f\ge0}$ is important: for example, $$y=\pm2x\kern.6em\not\kern-.6em\implies|y|=2x.$$

In your given example, $f=5e^{2x-2}\ge0.$

2 Answers2