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Math Help - need help, logarithms

  1. #1
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    Question 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)^(x) < 5^(-12-4x)
    4: ln(x)-ln(x-4) = 4
    5: ln(1-x) < 0

    how far i could get by myself:
    1: e^(x -2x) = e
    x-2x = 1
    x -2x -1 = 0
    abc formula:
    2+-(-2-4*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)^(x) < 5^(-12-4x)
    5^(-x) < 5^(-12-4x)
    -x < -12-4x
    0 < x-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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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stekemrt View Post
    what is x?
    1: e^(x*(x-2)) = e

    how far i could get by myself:
    1: e^(x -2x) = e
    x-2x = 1
    x -2x -1 = 0
    abc formula:
    2+-(-2-4*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?
    Look at your quadratic formula:
    \frac{2 \pm ((-2)^2 - 4 \cdot 1 \cdot (-1) )^{0.5}}{2}

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

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

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

    = 1 \pm 2^{0.5}

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stekemrt View Post
    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 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
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  4. #4
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    thx man, do you want a free ad on Webradioo: Internetradio, online radio luisteren. for a month ?
    thx again
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stekemrt View Post
    3: (1/5)^(x) < 5^(-12-4x)
    5^(-x) < 5^(-12-4x)
    -x < -12-4x
    0 < x-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: ( -\infty, -2),~(-2, 6),~(6, \infty).
    So for which interval(s) does your inequality hold true?

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stekemrt View Post
    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:
    \frac{x}{x - 4} = 2

    x = 2(x - 4)

    x = 2x - 8

    8 = 2x - x

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

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stekemrt View Post
    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
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  8. #8
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    Quote Originally Posted by topsquark View Post
    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:
    \frac{x}{x - 4} = 2

    x = 2(x - 4)

    x = 2x - 8

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

    -Dan
    hmm
    -2 < x
    x > 6

    right ?
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stekemrt View Post
    3: (1/5)^(x) < 5^(-12-4x)
    5^(-x) < 5^(-12-4x)
    -x < -12-4x
    0 < x-4x-12
    (x-6) (x+2)
    x=6 v x=-2
    Quote Originally Posted by stekemrt View Post
    hmm
    -2 < x
    x > 6

    right ?
    Plug representative values for each interval into the original inequality.
    -\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 < \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
    Last edited by topsquark; November 11th 2007 at 09:39 PM.
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  11. #11
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    Complete Solution

    Quote Originally Posted by topsquark View Post
    You have your test intervals: ( -\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.
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