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..
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thx again
Plug representative values for each interval into the original inequality.
< x < -2: Pick a useful number, say -1000000. This one appears to work. So x < -2 is in.
-2 < x < 6: Pick a representative again, say x = 0. This one doesn't work.
6 < x < : Pick x = 1000000. This also appears to work. So x > 6.
Thus the solution set is {x < -2} and {x > 6}.
(By the way, your version of the solution set is simply x > -2 since the two sets overlap.)
-Dan
Is simply, only inequation... you have to put the sign starting at + because the limit at [tex](+\infty) is positive (+)
___+_________-____________+____
-2 6
if the inequation is <0 you choose the - signus and if the inequation is > 0 you choose the + signus. In this example the inequation is < 0 then the solution is:
-2<x<6 or in other words x>-2 and x<6. If the inequation is >0 then the solution will be x<-2 or x>6.
I hope that this will be clear for your entire solution.
Best regards.