Results 1 to 4 of 4

Thread: inequality with absolute values

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    6

    inequality with absolute values

    I have the following inequality
    $\displaystyle \frac{1}{2}e^{-|x|} > \frac{1}{4}e^{-\frac{|x-1|}{2}}$
    By taking the log of both sides and some algebra, I get
    $\displaystyle -2|x|-2log2 > -|x-1|-2log4$
    At this point, I'm not sure what to do, since there's absolute values on both sides of the inequality, as well as other terms outside the absolute value. Any ideas? I know the approximate range is -2.4 < x < .8 . See the attached plot of the two functions.
    Attached Thumbnails Attached Thumbnails inequality with absolute values-hw3p1c.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Feb 2009
    From
    Posts
    79
    Quote Originally Posted by r2d2bol View Post
    I have the following inequality
    $\displaystyle \frac{1}{2}e^{-|x|} > \frac{1}{4}e^{-\frac{|x-1|}{2}}$
    By taking the log of both sides and some algebra, I get
    $\displaystyle -2|x|-2log2 > -|x-1|-2log4$
    At this point, I'm not sure what to do, since there's absolute values on both sides of the inequality, as well as other terms outside the absolute value. Any ideas? I know the approximate range is -2.4 < x < .8 . See the attached plot of the two functions.
    I believe your taking the log step is incorrect.

    To solve this inequality it may be useful to do the following.

    $\displaystyle 4* (\frac{1}{2}e^{-|x|} > \frac{1}{4}e^{-\frac{|x-1|}{2}})$


    $\displaystyle 2e^{-|x|} > e^{-\frac{|x-1|}{2}}$

    Dividing both sides by $\displaystyle e^{-|x|} $

    $\displaystyle 2 > \frac{ e^{-\frac{|x-1|}{2}} }{e^{-|x|} }$

    $\displaystyle 2 > e^{\frac{-|x-1|}{2} + |x| }$


    $\displaystyle 2 > e^{\frac{-|x-1|+2|x|}{2}}$

    Solve the inequality for x >= 1

    and for x< 1

    And then viola.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    6
    Ok, starting from here
    $\displaystyle 2 > e^{\frac{-|x-1|+2|x|}{2}}$
    take the log and then multiplying times 2
    $\displaystyle 2log2 >-|x-1| + 2|x|$
    I think the answer I need is:
    $\displaystyle
    2log2 <x-1 -2x$
    $\displaystyle 2log2+1<-x$
    $\displaystyle -2.3863 < x
    $
    and
    $\displaystyle 2log2>x-1+2x$
    $\displaystyle 2log2+1>3x$
    $\displaystyle 2log2+1>3x$
    $\displaystyle .7954 > x$

    but I don't really understand why this works out this way.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Feb 2009
    From
    Posts
    79
    It's just mathematics
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Sep 7th 2011, 07:11 AM
  2. [SOLVED] Inequality with absolute values
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 27th 2010, 03:35 PM
  3. Inequality with absolute values
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Nov 15th 2010, 08:35 AM
  4. sketch an inequality with absolute values?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 23rd 2009, 06:02 PM
  5. Replies: 5
    Last Post: Sep 6th 2008, 08:38 PM

Search Tags


/mathhelpforum @mathhelpforum