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Isolate y from e

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  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    X=e^y i just need to do that i think this is maybe log but i dont know



    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 5, 2014 #2

    LCKurtz

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    Show us what happens if you take the natural logarithm of both sides.
     
  4. Oct 5, 2014 #3
    Log(x) =ylog(e)???
     
  5. Oct 5, 2014 #4

    SteamKing

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    But what is the natural log of e? Think about what a natural logarithm is, and how it's defined.
     
  6. Oct 5, 2014 #5
    Its one, can you apoint me to the solution i am lost
     
  7. Oct 5, 2014 #6

    Mark44

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  8. Oct 5, 2014 #7
    I found the solution y=-log(x)
     
  9. Oct 5, 2014 #8
    Sorry its y=-lnx
     
  10. Oct 5, 2014 #9

    Mark44

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    Neither one of these is correct. Please show what you did to get your last equation.
     
  11. Oct 5, 2014 #10
    X=e^-y then lnx=-ylne then (lnx=-y)-1.... Then (y=-lnx) or y=ln(1/x)
     
  12. Oct 5, 2014 #11

    Matterwave

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    Your original post had the equation x=e^y rather than x=e^-y. Which problem are you doing? The solution you have here is correct for x=e^-y except I'm not sure why you have a -1 in the 3rd equation which disappears in later equations yielding the correct answer.
     
  13. Oct 5, 2014 #12

    SteamKing

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    If log (e) = 1, then what is y log (e)?
     
  14. Oct 5, 2014 #13
    Sorry i forgot the - and the-1 its just to change the - x to x and i was wrong whit log its ln
     
  15. Oct 5, 2014 #14

    Mark44

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    If the original equation was supposed to be x = e-y, then the equivalent equation is y = -ln(x). By "equivalent" I mean that every ordered pair (x, y) that satisfies the first equation also satisfies the second equation.
     
  16. Oct 5, 2014 #15
    Yes thanks dude, i really apreciate the help
     
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