URGENT: SAC next period!!

[Apink!]

Sorry guys,
I have a quick question. (I have a SAC next period!)
Solve -1< 2- (1/2)e^(1+x)/2 <1

Arrggg
Apparantly the correct answer is (2ln(2)-1, infinity) U (-infinity, 2ln(6)-1)
I think the answer should be the intersection of these domains!!

Could someone please help me out??

[wunderkind52]

Sorry guys,
I have a quick question. (I have a SAC next period!)
Solve -1< 2- (1/2)e^(1+x)/2 <1

Arrggg
Apparantly the correct answer is (2ln(2)-1, infinity) U (-infinity, 2ln(6)-1)
I think the answer should be the intersection of these domains!!

Could someone please help me out??

Always sketch the graphs! Sorry I'm in class so can't verify the answer

[Apink!]

I did, and I think I'm right
https://www.wolframalpha.com/input/?i=-1%3C2-%281%2F2%29e%5E%281%2Bx%29%2F2%3C1+solve+for+x
wolfram think so too, I think...

[Apink!]

I also don't understand where the modulus sign came from

[heids]

I also don't understand where the modulus sign came from

Remember, log functions can't have negative numbers in the brackets; when there was a squared sign, no matter what x is, the stuff in the brackets is positive.  But when you took the squared sign away, you have to put in mod signs to make sure it's still positive.
(Arggh, this is phrased terribly, can't believe how utterly rusty my maths has become!!)

[Apink!]

Thank you wunderkind and bangali_lok!
Turns out those questions weren't on the SAC!
But thank you!!