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Math Help - inequality with absolute values

  1. #1
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    inequality with absolute values

    I have the following inequality
    \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
    -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  
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  2. #2
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    Quote Originally Posted by r2d2bol View Post
    I have the following inequality
    \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
    -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.

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


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

    Dividing both sides by e^{-|x|}

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

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


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

    Solve the inequality for x >= 1

    and for x< 1

    And then viola.
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  3. #3
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    Ok, starting from here
     2 >  e^{\frac{-|x-1|+2|x|}{2}}
    take the log and then multiplying times 2
    2log2 >-|x-1| + 2|x|
    I think the answer I need is:
    <br />
2log2 <x-1 -2x
    2log2+1<-x
    -2.3863 < x<br />
    and
    2log2>x-1+2x
    2log2+1>3x
    2log2+1>3x
    .7954 > x

    but I don't really understand why this works out this way.
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  4. #4
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    It's just mathematics
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