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moongazer Group Title

Find the value of x and y that will satisfy the system

  • 2 years ago
  • 2 years ago

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  1. moongazer Group Title
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    e^x - y = -2 e^x + y = 1

    • 2 years ago
  2. Matrix65 Group Title
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    is x-y and x+y both exponents

    • 2 years ago
  3. sauravshakya Group Title
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    |dw:1349690570593:dw|

    • 2 years ago
  4. sauravshakya Group Title
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    Is that the question?

    • 2 years ago
  5. Matrix65 Group Title
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    introduce In both sides

    • 2 years ago
  6. sauravshakya Group Title
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    If that is the question then it has no real roots.

    • 2 years ago
  7. moongazer Group Title
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    how do I solve it?

    • 2 years ago
  8. moongazer Group Title
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    (e^x) - y = -2 (e^x) + y = 1

    • 2 years ago
  9. moongazer Group Title
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    sorry for the late reply

    • 2 years ago
  10. sauravshakya Group Title
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    Then Add equation one and two ........

    • 2 years ago
  11. prawal Group Title
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    i think no real solution... u might get5 imaginary....

    • 2 years ago
  12. hartnn Group Title
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    still e^x = -1/2 has no real solution for x

    • 2 years ago
  13. sauravshakya Group Title
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    Yes still no real solution

    • 2 years ago
  14. hartnn Group Title
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    y comes out to be 3/2

    • 2 years ago
  15. hartnn Group Title
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    u have choices/options ?

    • 2 years ago
  16. sauravshakya Group Title
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    But that will not be true.

    • 2 years ago
  17. moongazer Group Title
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    I don't have

    • 2 years ago
  18. FaisalAyaz Group Title
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    imaginary solution is how?

    • 2 years ago
  19. sauravshakya Group Title
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    |dw:1349691346340:dw|

    • 2 years ago
  20. sauravshakya Group Title
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    |dw:1349691384296:dw|

    • 2 years ago
  21. moongazer Group Title
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    that's also what I got

    • 2 years ago
  22. hartnn Group Title
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    http://www.wolframalpha.com/input/?i=e%5Ex+%3D+-1%2F2

    • 2 years ago
  23. FaisalAyaz Group Title
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    ok

    • 2 years ago
  24. moongazer Group Title
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    Isn't it that e^(ln -1/2) =-1/2 ?

    • 2 years ago
  25. moongazer Group Title
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    @hartnn @sauravshakya

    • 2 years ago
  26. hartnn Group Title
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    yes, thats correct but y did u caculate that ?

    • 2 years ago
  27. moongazer Group Title
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    nope

    • 2 years ago
  28. hartnn Group Title
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    i meant yes, thats correct but why did u caculate that ?

    • 2 years ago
  29. prawal Group Title
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    e^(ln -1/2) =-1/2 this is hell wrong... !! the equation is not defined...

    • 2 years ago
  30. moongazer Group Title
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    to find y

    • 2 years ago
  31. moongazer Group Title
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    I got y=3/2

    • 2 years ago
  32. hartnn Group Title
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    yes, there are no real solutions to this.....

    • 2 years ago
  33. prawal Group Title
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    impossible man..

    • 2 years ago
  34. moongazer Group Title
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    What should I write on my paper?

    • 2 years ago
  35. hartnn Group Title
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    i think u just write e^x=-1/2, y =3/2 and leave it

    • 2 years ago
  36. prawal Group Title
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    u can do the above thing or write no real solution...

    • 2 years ago
  37. hartnn Group Title
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    *and write no real solution

    • 2 years ago
  38. moongazer Group Title
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    I think I got the answer

    • 2 years ago
  39. jhonyy9 Group Title
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    ln of a negativ number - how it is possible ? --- little up check it please !!!

    • 2 years ago
  40. jhonyy9 Group Title
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    @hartnn ,@sauravshakya

    • 2 years ago
  41. moongazer Group Title
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    I got x = ln -1 and y = 1

    • 2 years ago
  42. jhonyy9 Group Title
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    @sauravshakya

    • 2 years ago
  43. moongazer Group Title
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    I think that is the answer

    • 2 years ago
  44. jhonyy9 Group Title
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    check where is defined the logarithm funvtion please

    • 2 years ago
  45. jhonyy9 Group Title
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    function --- sorry

    • 2 years ago
  46. moongazer Group Title
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    I forgot to add e^x

    • 2 years ago
  47. jhonyy9 Group Title
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    e^x - y = -2 e^x + y = 1

    • 2 years ago
  48. hartnn Group Title
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    ln(-1), 1 does not satisfy 2nd equation ln(-1) is imaginary = i\(\pi\)

    • 2 years ago
  49. jhonyy9 Group Title
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    ln(-1) how do you mean this ?

    • 2 years ago
  50. moongazer Group Title
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    it should be ln -1/2 =x and 3/2=y

    • 2 years ago
  51. jhonyy9 Group Title
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    when by definition logb X so x need being allways positiv

    • 2 years ago
  52. hartnn Group Title
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    \(e^{i\pi}=-1\) by euler's relation

    • 2 years ago
  53. jhonyy9 Group Title
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    e^x - y = -2 e^x + y = 1 so and for these equations ?

    • 2 years ago
  54. hartnn Group Title
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    but if x is imaginary, y must be imaginary also, right ?

    • 2 years ago
  55. jhonyy9 Group Title
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    but in this case i think that is very simple us ln to both sides for every two equations and will get the right answers

    • 2 years ago
  56. moongazer Group Title
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    when you substitute x with ln -1/2 it will be e^(ln -1/2) - y = -2 then -1/2-y=-2 y=3/2

    • 2 years ago
  57. moongazer Group Title
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    I think my teacher doesn't need to know the decimal value of x.

    • 2 years ago
  58. hartnn Group Title
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    ok, then write that....

    • 2 years ago
  59. moongazer Group Title
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    ok, thanks for the answers :)

    • 2 years ago
  60. moongazer Group Title
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    why is it ln (negative number) is undefined?

    • 2 years ago
  61. jhonyy9 Group Title
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    by domain definition of logarithms

    • 2 years ago
  62. jhonyy9 Group Title
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    you have learned it sure --- i think ,yes ?

    • 2 years ago
  63. moongazer Group Title
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    ohh yes

    • 2 years ago
  64. moongazer Group Title
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    because of the graph

    • 2 years ago
  65. jhonyy9 Group Title
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    good so than all is understood it right good luck bye

    • 2 years ago
  66. moongazer Group Title
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    Thanks :) do you know how to decompose a function?

    • 2 years ago
  67. jhonyy9 Group Title
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    how do you think it ?

    • 2 years ago
  68. moongazer Group Title
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    partial fractions the one with cases. I got an odd looking expression

    • 2 years ago
  69. moongazer Group Title
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    i'll post it

    • 2 years ago
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is replying to Can someone tell me what button the professor is hitting...

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