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moongazer

  • 3 years ago

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

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  1. moongazer
    • 3 years ago
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    e^x - y = -2 e^x + y = 1

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

  3. sauravshakya
    • 3 years ago
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    |dw:1349690570593:dw|

  4. sauravshakya
    • 3 years ago
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    Is that the question?

  5. Matrix65
    • 3 years ago
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    introduce In both sides

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

  7. moongazer
    • 3 years ago
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    how do I solve it?

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

  9. moongazer
    • 3 years ago
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    sorry for the late reply

  10. sauravshakya
    • 3 years ago
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    Then Add equation one and two ........

  11. prawal
    • 3 years ago
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    i think no real solution... u might get5 imaginary....

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

  13. sauravshakya
    • 3 years ago
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    Yes still no real solution

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

  15. hartnn
    • 3 years ago
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    u have choices/options ?

  16. sauravshakya
    • 3 years ago
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    But that will not be true.

  17. moongazer
    • 3 years ago
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    I don't have

  18. FaisalAyaz
    • 3 years ago
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    imaginary solution is how?

  19. sauravshakya
    • 3 years ago
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    |dw:1349691346340:dw|

  20. sauravshakya
    • 3 years ago
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    |dw:1349691384296:dw|

  21. moongazer
    • 3 years ago
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    that's also what I got

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

  23. FaisalAyaz
    • 3 years ago
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    ok

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

  25. moongazer
    • 3 years ago
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    @hartnn @sauravshakya

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

  27. moongazer
    • 3 years ago
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    nope

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

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

  30. moongazer
    • 3 years ago
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    to find y

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

  32. hartnn
    • 3 years ago
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    yes, there are no real solutions to this.....

  33. prawal
    • 3 years ago
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    impossible man..

  34. moongazer
    • 3 years ago
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    What should I write on my paper?

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

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

  37. hartnn
    • 3 years ago
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    *and write no real solution

  38. moongazer
    • 3 years ago
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    I think I got the answer

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

  40. jhonyy9
    • 3 years ago
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    @hartnn ,@sauravshakya

  41. moongazer
    • 3 years ago
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    I got x = ln -1 and y = 1

  42. jhonyy9
    • 3 years ago
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    @sauravshakya

  43. moongazer
    • 3 years ago
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    I think that is the answer

  44. jhonyy9
    • 3 years ago
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    check where is defined the logarithm funvtion please

  45. jhonyy9
    • 3 years ago
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    function --- sorry

  46. moongazer
    • 3 years ago
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    I forgot to add e^x

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

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

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

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

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

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

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

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

  55. jhonyy9
    • 3 years ago
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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

  56. moongazer
    • 3 years ago
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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

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

  58. hartnn
    • 3 years ago
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    ok, then write that....

  59. moongazer
    • 3 years ago
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    ok, thanks for the answers :)

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

  61. jhonyy9
    • 3 years ago
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    by domain definition of logarithms

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

  63. moongazer
    • 3 years ago
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    ohh yes

  64. moongazer
    • 3 years ago
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    because of the graph

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

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

  67. jhonyy9
    • 3 years ago
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    how do you think it ?

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

  69. moongazer
    • 3 years ago
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    i'll post it

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