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I found the answers, but the back of the book has something different?

  1. May 24, 2009 #1
    1. The problem statement, all variables and given/known data

    The problem is

    ln ( 2 - 5x) > 2

    2. Relevant equations

    The answers for the back of the book is x < (2-e^2) / 5

    3. The attempt at a solution

    Here's my attempt

    ln ( 2 - 5x ) > 2

    = ln 2 - ln 5x > 2

    = - ln 5x > 2 - ln 2

    = x < (2- ln 2) / -ln 5

    Is there a way I can simplify this even more guys. thanks for all your help so far
     
  2. jcsd
  3. May 24, 2009 #2

    danago

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    Have a look at that line ive bolded. Perhaps have a look at your logarithm rules again, and see if you can see what you have done wrong. The answer the book has is infact correct.
     
  4. May 24, 2009 #3
    Thank you so much for over looking my work.

    Let's see, I've think i've simplied it correctly this time, but is there a way to get it really bone dried simplied? Thanks for your help

    I've got

    = e^2 > 2- 5x

    = e^2 - 2 > -5 x

    = ( e^2 - 2) / -5 < x

    I can't get the signs to match for some reason and negative and positive stuff. I'm so close my good sir
     
  5. May 24, 2009 #4

    danago

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    Your answer is correct now. It is exactly the same as the books answer, but they have just multiplied both the top and bottom of the fraction by -1, just to make it look a bit neater.

    It is basically the same idea as -5/-2 = 5/2
     
  6. May 24, 2009 #5
    You're correct sir, but I still can't get my greater than or less than sign to match up. i'm mulitplying top and bottom by -1 so that's twice making my < sign stay the same. Thanks for over looking my work my kind sir, but just missing one part of the puzzle : )
     
  7. May 24, 2009 #6

    danago

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    Oh sorry didnt even realise you had the inequality sign the wrong way around. How did you you go from ln (2-5x) > 2 to e^2 > 2- 5x? It seems you have got the inequality sign the wrong way round here.
     
  8. May 24, 2009 #7
    I dont understand, i wrote down the problem correctly, and i just change the original problem to better see it to:

    log (base e) (2-5x ) > 2

    Then I change it to the exponential form

    e^2 > 2-5x

    Did I miss something with the change of signs?
     
  9. May 24, 2009 #8

    danago

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    If ln (2-5x) > 2, then that should imply that 2-5x > e^2, not the other way around.
     
  10. May 24, 2009 #9
    But I didn't write down the problem wrong. I really don't understand where I missed the sign change.
     
  11. May 24, 2009 #10

    danago

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    You went from ln (2-5x) > 2 to e^2 > 2-5x. All im saying is that you got the inequality sign the wrong way around in that first step.
     
  12. May 24, 2009 #11
    I see I have to keep an eye on that, if I'm having the same confuzino here. BTW, thank you for your time, can't say that enough, im stressing about a big math final.

    Like for example on antoher one im stuck

    e ^( x^2 - 4x) > = e ^ 5

    I take the natural log of both sides and it becomes

    X^2 - 4x >= 5

    Then I subtract 5

    x^2 - 4x - 5

    and I factor

    (x - 5 ) ( x + 1 ) >= 0

    so x => 5 or x >= -1

    But the back of the book says x <= -1 or x >=5

    These signs are killing me, my friend
     
  13. May 24, 2009 #12

    danago

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    All good up until (x - 5 ) ( x + 1 ) >= 0.

    What we are essentially saying there is that the product of two numbers (x-5) and (x+1) is positive. That means they must be the same sign; either both positive, or both negative.

    Lets look at the case where they are both positive:
    x-5>0 Therefore x>5
    x+1>0 Therefore x>-1

    We need when they are BOTH positive, so that only occurs when x>5.

    If you do the same and find when they are BOTH negative, you will see that x<-1.

    Does that make sense to you?
     
  14. May 24, 2009 #13
    danago, Thank you so much! yes that does make sense to me, because a positive times a positive equals a positive and a negative times a negative is a positive. I see why that would be a solutkion

    Is there anyway I can repay you for all your help?
     
  15. May 24, 2009 #14

    danago

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    Exactly right :)

    Just send the cheque to... haha nah no need to repay me in any way, i like helping out when i can. Good luck with your final!
     
  16. May 24, 2009 #15
    Can't thank you enough, but i'll be sure to help other members in return with what i can. Thanks again
     
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