Is My Calculation Correct for O.4^(|x-2|/|x+2|) = 0.4^2?

[Plutonium88]

O.4^(|x-2|/|x+2|) = 0.4^2 Help Me :(

O.4^(|x-2|/|x+2|) < 0.4^2

so (|x-2|/|x+2|) < 2

restrictions: x≠-2

so for:

|x-2|------> (x-2) when x ≥ 2
-------> -(x-2) when x<2

|x+2| -------> (x+2) when x≥ -2
----------> -(x+2) when x<-2
with this i now have three separate intervals for calculation

(1)X< -2 (2)-2<x<2 (3)x>2

where-2, and 2 are holes

so for the ineterval x<-2

-(x-2)/-(x+2) < 2

negatives cancel multipy by (x+2) on both sides

x-2 < 2(x+2)
x - 2 < 2x+4
x> -6
this fits within its domain.
--------------------------------------------------------------------------------
for interval -2<x<2

(x-2)/(x+2) < 2

-x+2 < 2x +4
-2 < 3x
x> -2/3

this fits within its domain
-----------------------------------------------------------------
interval X>2

(x-2)/(x+2) < 2
x-2 < 2x + 4
x>-6can some one tell me if my intervals are correct?
wow i can't believe i spent my time on this stupid problem all day, just because in my scond interval. i forgot to switch the inequality sign in divisoninteval 2 -2<x<2

-(x-2)< 2(x+2)
-x + 2 < 2x +4
-2< 3x
DIVIDE siwtch sign :'"""""""(

x < -2/3there for since x> -6 for interval x<-2
and x< -2/3 for -2<x<2

xE(-6, -2)U(-2, -2/3)

 

[pcm]

O.4^(|x-2|/|x+2|) < 0.4^2

so (|x-2|/|x+2|) < 2

restrictions: x≠-2

a^x is a decreasing functions if 0<a<1; so 0.4^f(x) is decreasing, therefore, (|x-2|/|x+2|)>2;

 

[Plutonium88]

Actually what I said is wrong di dividing does not change inequalityNmy answer is wrong please hel

 

[Bohrok]

(|x-2|/|x+2|) < 2 → |x-2| < 2|x+2|

I think the easiest way to solve this would be by breaking |x| < y into the two cases x < y and x > -y for positive y, then apply that again since the original inequality has to absolute values. This will ultimately give you four cases to work out, and each of the two solutions will be repeated.

 

[pcm]

so for the ineterval x<-2

-(x-2)/-(x+2) < 2

negatives cancel multipy by (x+2) on both sides

x-2 < 2(x+2)
x - 2 < 2x+4
x> -6
this fits within its domain.
--------------------------------------------------------------------------------

x<-2 ,implies x+2<0
so when multiplying by x+2 ,inequality sign will change.

 

[Plutonium88]

x<-2 ,implies x+2<0
so when multiplying by x+2 ,inequality sign will change.

ty for trying to help, i just got it on friday.what you have to do is solve the inequality like this for the three different intervals.
0.4^|x-2|/|x+2| < 0.4^2
|x-2|/|x+2| > 2
|x-2|/|x+2| > 2

|x-2|/|x+2| - 2 > 0
(|x-2|-2|x+2|)/|x+2| > 0then when you solve like this, you can do interval charts for each of the domains/intervals and find your answers. ty for the help tho. A friend at school helped me solve it, so respects to him also.