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Solving this inequation

  1. Feb 9, 2008 #1
    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. jcsd
  3. Feb 9, 2008 #2
    Your step 2. is wrong I believe. Logically, EITHER [tex]x^2 - x - 6 < 0[/tex] or [tex]log(abs(x^2-4))<0[/tex]. What you meant to use is exclusive OR, [tex]\oplus[/tex], which is defined as (P AND ~Q) OR (~P AND Q).
     
    Last edited: Feb 9, 2008
  4. Feb 9, 2008 #3
    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.
     
  5. Feb 9, 2008 #4
    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
     
  6. Feb 9, 2008 #5
    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.
     
  7. Feb 9, 2008 #6
    Like in this video? http://youtube.com/watch?v=pV3cZ1zbuvs
     
  8. Feb 9, 2008 #7
    I was thinking that the only regions that will solve that were going to come from when one multiplicand or the other - but not both - were less than zero. The OP's work makes no attempt to confirm that the other multiplicand is positive when one is negative. Your suggestion is probably easier to implement though.
     
    Last edited: Feb 9, 2008
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