Please, can someone help me to solve this:
|x - 2| = 1/e
I really don't know the way in which I could solve an absolute value.

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    $\begingroup$ Either $x-2={1\over e}$ or $-(x-2)={1\over e}$. Can you take it from here? $\endgroup$ – David Mitra Dec 14 '11 at 12:13
  • $\begingroup$ So, I have to join the values of the two cases (if it's positive, or if it's negative) and take the solution of the join? $\endgroup$ – Overflowh Dec 14 '11 at 12:18
  • $\begingroup$ You would solve each separately. The original equation will have two solutions. So, yes "join" makes sense. The solution set of the original equation is the union (join) of the solution sets of the equations from the two cases. $\endgroup$ – David Mitra Dec 14 '11 at 12:23
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    $\begingroup$ It might also help you if you draw a graph: wolframalpha.com/input/?i=plot+%7B%7Cx-2%7C%2C+1%2Fe%7D $\endgroup$ – Martin Sleziak Dec 14 '11 at 12:27
  • $\begingroup$ For the intuition, it is useful to know that $|x-2|$ is the distance between $x$ and $2$. This distance is $1/e$ precisely when $x=2+1/e$ and when $x=2-1/e$, that is, $1/e$ to the right and $1/e$ to the left of $2$. (I am assuming that by $e$ you mean the base for the natural logarithms.) $\endgroup$ – André Nicolas Dec 14 '11 at 17:23

From the definition of absolute value, $x$ is a solution of $$\tag{1}|x-2|={1\over e}$$ if and only if $$\tag{2}x-2={1\over e}$$ or $$\tag{3}-(x-2)={1\over e}.$$ (Since $|x-2|$ is either $x-2$ or $-(x-2)\,$.)

For emphasis: if $a$ is a solution of either equation (2) or equation (3), it must be a solution of equation (1). On the other hand, if $a$ is a solution of equation (1), it must be a solution of one of equations (2) or (3).

So, the solution set of equation (1) is precisely the union of the solution sets to equations (1) and (2).

You need to solve both of equations (1) and (2) (which I leave to you). Then you can state that the solutions to equation (1) are all the solutions found in the previous sentence.


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