1. ## need help, logarithms

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..

2. Originally Posted by stekemrt
what is x?
1: e^(x*(x-2)) = e

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?
$\displaystyle \frac{2 \pm ((-2)^2 - 4 \cdot 1 \cdot (-1) )^{0.5}}{2}$

$\displaystyle = \frac{2 \pm ((4 + 4 )^{0.5}}{2}$

$\displaystyle = \frac{2 \pm 8^{0.5}}{2}$

$\displaystyle = \frac{2 \pm 2 \cdot 2^{0.5}}{2}$

$\displaystyle = 1 \pm 2^{0.5}$

-Dan

3. Originally Posted by stekemrt
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 ?
Your condition simplifies to 1 > -1. This is a true statement, but says nothing of value about your problem.

The point is that $\displaystyle e^x > e^{-x}$ only when x is positive. If x is negative (as in my example above) then this is no longer true.

-Dan

thx again

5. Originally Posted by stekemrt
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?
You have your test intervals: $\displaystyle ( -\infty, -2),~(-2, 6),~(6, \infty)$.
So for which interval(s) does your inequality hold true?

-Dan

6. Originally Posted by stekemrt
4x)
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
You're doing great! You've got a minor sign error in that last line, though. Let's see how you would solve this one:
$\displaystyle \frac{x}{x - 4} = 2$

$\displaystyle x = 2(x - 4)$

$\displaystyle x = 2x - 8$

$\displaystyle 8 = 2x - x$

How do you finish this? The numbers are different, but the solution method is exactly the same.

-Dan

7. Originally Posted by stekemrt
5: ln(1-x) < 0

1-x < 1
0 < x
but according to the book the answer is 1>x>0
help..
Look at your original expression. What happens if x > 1? Your solution method is fine.

-Dan

8. Originally Posted by topsquark
You're doing great! You've got a minor sign error in that last line, though. Let's see how you would solve this one:
$\displaystyle \frac{x}{x - 4} = 2$

$\displaystyle x = 2(x - 4)$

$\displaystyle x = 2x - 8$

$\displaystyle 8 = 2x - x$

How do you finish this? The numbers are different, but the solution method is exactly the same.

-Dan
hmm
1) 8 = 2x - x
8=x

2) x = 2x - 8
8=x

3) x = 2(x - 4)
x=2x - 8
8=x

4) .. let me think for 1 more minute..

EDIT:
x=2(x-4)
x=2x - 8
8=x
rofl,, i just wanted to exercise some..
math is quite cool ^^

9. Originally Posted by topsquark
You have your test intervals: $\displaystyle ( -\infty, -2),~(-2, 6),~(6, \infty)$.
So for which interval(s) does your inequality hold true?

-Dan
hmm
-2 < x
x > 6

right ?

10. Originally Posted by stekemrt
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
Originally Posted by stekemrt
hmm
-2 < x
x > 6

right ?
Plug representative values for each interval into the original inequality.
$\displaystyle -\infty$ < 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 < $\displaystyle \infty$: 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

11. ## Complete Solution

Originally Posted by topsquark
You have your test intervals: $\displaystyle ( -\infty, -2),~(-2, 6),~(6, \infty)$.
So for which interval(s) does your inequality hold true?

-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.