I've got a problem with some math exercises

* I cant use a calculator, i must answer like this for example: x = ln(3)

what is x?

1: e^(x*(x-2)) = e

2: e^x > e^-x

3: (1/5)^(x²) < 5^(-12-4x)

4: ln(x)-ln(x-4) = 4

5: ln(1-x) < 0

how far i could get by myself:

1: e^(x² -2x) = e

x²-2x = 1

x² -2x -1 = 0

abc formula:

2+-(-2²-4*1*-1)^0.5

2

x=2,41 v x=0,41

I cant do this out of my head, according to the book the answer was 1+- (2)^0.5, how did they do it?

2:e^x > e^-x

I thought this was simply x > -x , but it turned out to be x > 0 why is it x > 0? and was my answer correct also ?

3: (1/5)^(x²) < 5^(-12-4x)

5^(-x²) < 5^(-12-4x)

-x² < -12-4x

0 < x²-4x-12

(x-6) (x+2)

x=6 v x=-2

ok im not sure if its x<6 or x>6, or x<-2 or x>-2. Im always doing this wrong, could somebody explain how it works?

4: ln(x)-ln(x-4) = 4

ln(x/(x-4)) = 4

x/(x-4) = e^4

x=e^4*(x-4)

x=e^4x-4e^4

4e^4=e^4x+x

and thats how far i could get

5: ln(1-x) < 0

1-x < 1

0 < x

but according to the book the answer is 1>x>0

help..