inequality with absolute values

**r2d2bol** · May 6th 2009, 06:59 PM

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.

**Banned for attempted hacking** · May 6th 2009, 08:09 PM

Originally Posted by r2d2bol

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.

**r2d2bol** · May 6th 2009, 09:09 PM

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.

**Banned for attempted hacking** · May 7th 2009, 06:39 AM

It's just mathematics

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