# Thread: inequality with absolute values

1. ## 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.

2. Originally Posted by r2d2bol
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.

3. 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.

4. It's just mathematics