# Solving this inequation

1. Feb 9, 2008

### fishingspree2

Hello,

abs(x² - 4)^(x² - x - 6) < 1

here's my work:

1. taking the log of both sides (can i do this even if it's an inequality?)
log (abs(x² - 4)^(x² - x - 6)) < log 1
which is
(x² - x - 6)*(log(abs(x² - 4))) < 0

2. I can then say:
(x² - x - 6) < 0
OR
log(abs(x² - 4)) < 0

3. Solving:
(x² - x - 6) < 0
gives: -2 < x < 3

log(abs(x² - 4)) < 0
gives: -sqrt(5) < x < -sqrt(3) and sqrt(3) < x < sqrt(5)

so the solution set is:
]-√5 , -2[ U ]√5 , 3[ U ]-√3,√3[

Questions:
1. Is my work correct?
2. How would you do it, and can you please show work?

Thank you

EDIT: It's not really a homework... I was just wondering how would you guys do it :S
Sorry

2. Feb 9, 2008

### jhicks

Your step 2. is wrong I believe. Logically, EITHER $$x^2 - x - 6 < 0$$ or $$log(abs(x^2-4))<0$$. What you meant to use is exclusive OR, $$\oplus$$, which is defined as (P AND ~Q) OR (~P AND Q).

Last edited: Feb 9, 2008
3. Feb 9, 2008

### John Creighto

Why not just test it numerically. See if a point on the boundary gives you the value 1. See if a point inside of the boundary gives you a value less then one. See if a point outside the boundary gives you a value greater then one.

4. Feb 9, 2008

### fishingspree2

Sorry, my maths skills aren't that advanced.

What is the name of the crosshair symbol? Thank you

Also, I'd like to know some other ways to do it

5. Feb 9, 2008

### John Creighto

Really? How do we know they both aren't less then zero? Does it really matter anyway. I don't know why the original poster bothered keeping the inequality signs. I would solve the problem by finding all possible intersections and then numerically testing to see which side of the boundaries the region of interest lies in.

6. Feb 9, 2008