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Ln(x-2) = -2+ln(x)

  1. Aug 30, 2007 #1
    I have no idea where to start. I need to know how solve this.

    I graphed both of the equations and then found the intersection, but it is a non repeating decimal and my answer has to be exact. I don't know how else to do it.

    Please HELP!
     
  2. jcsd
  3. Aug 30, 2007 #2
    any work? gotta show some b4 we can help.

    and exactly what are you solving for?
     
  4. Aug 30, 2007 #3
    Hm..without giving you the answer.

    Try to get rid of those logs, what can you do to get rid of the logs?

    If that doesn't work try combining the logs to get 1 single log and see if that gives you the answer.
     
  5. Aug 30, 2007 #4

    Dick

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    Take e to the power of both sides.
     
  6. Aug 30, 2007 #5
    you can raise e to both sides?
     
  7. Aug 30, 2007 #6
    then is it simply just solving for x?
     
  8. Aug 30, 2007 #7
    i know that e^(ln(x-2)) is x-2 but i don't know what happens to the other side
     
  9. Aug 30, 2007 #8

    Dick

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    I thought you knew you were solving for x. What else is there to solve for? And sure you can raise e to both sides. If a=b, then e^a=e^b.
     
  10. Aug 30, 2007 #9

    Dick

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    What's e^(a+b). And how are your replies showing up so fast?
     
  11. Aug 30, 2007 #10
    when you need help = constantly refreshing ;) and through e-mail notifications
     
  12. Aug 30, 2007 #11
    e^a * e^b ???
     
  13. Aug 30, 2007 #12

    Dick

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    My email notifications show up in a slug-like fashion. Like at least a minute or two between. I thought the forum was delaying for last minute edits. Guess I'm wrong, or the OP is anticipating the reply.
     
  14. Aug 30, 2007 #13

    Dick

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    You did it again. YES!
     
  15. Aug 30, 2007 #14
    oh dang.. i got it i think. -2/(e^(-2)-1) maybe?
     
  16. Aug 30, 2007 #15

    Dick

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    Yes!!!!!!! I know I've got to get this in quick.
     
  17. Aug 30, 2007 #16
    sweet.. thanks a lot! all i needed was a little push in the right direction haha
     
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