## moongazer 4 years ago Find the value of x and y that will satisfy the system

1. moongazer

e^x - y = -2 e^x + y = 1

2. anonymous

is x-y and x+y both exponents

3. anonymous

|dw:1349690570593:dw|

4. anonymous

Is that the question?

5. anonymous

introduce In both sides

6. anonymous

If that is the question then it has no real roots.

7. moongazer

how do I solve it?

8. moongazer

(e^x) - y = -2 (e^x) + y = 1

9. moongazer

10. anonymous

Then Add equation one and two ........

11. anonymous

i think no real solution... u might get5 imaginary....

12. hartnn

still e^x = -1/2 has no real solution for x

13. anonymous

Yes still no real solution

14. hartnn

y comes out to be 3/2

15. hartnn

u have choices/options ?

16. anonymous

But that will not be true.

17. moongazer

I don't have

18. anonymous

imaginary solution is how?

19. anonymous

|dw:1349691346340:dw|

20. anonymous

|dw:1349691384296:dw|

21. moongazer

that's also what I got

22. hartnn
23. anonymous

ok

24. moongazer

Isn't it that e^(ln -1/2) =-1/2 ?

25. moongazer

@hartnn @sauravshakya

26. hartnn

yes, thats correct but y did u caculate that ?

27. moongazer

nope

28. hartnn

i meant yes, thats correct but why did u caculate that ?

29. anonymous

e^(ln -1/2) =-1/2 this is hell wrong... !! the equation is not defined...

30. moongazer

to find y

31. moongazer

I got y=3/2

32. hartnn

yes, there are no real solutions to this.....

33. anonymous

impossible man..

34. moongazer

What should I write on my paper?

35. hartnn

i think u just write e^x=-1/2, y =3/2 and leave it

36. anonymous

u can do the above thing or write no real solution...

37. hartnn

*and write no real solution

38. moongazer

I think I got the answer

39. jhonyy9

ln of a negativ number - how it is possible ? --- little up check it please !!!

40. jhonyy9

@hartnn ,@sauravshakya

41. moongazer

I got x = ln -1 and y = 1

42. jhonyy9

@sauravshakya

43. moongazer

I think that is the answer

44. jhonyy9

check where is defined the logarithm funvtion please

45. jhonyy9

function --- sorry

46. moongazer

47. jhonyy9

e^x - y = -2 e^x + y = 1

48. hartnn

ln(-1), 1 does not satisfy 2nd equation ln(-1) is imaginary = i$$\pi$$

49. jhonyy9

ln(-1) how do you mean this ?

50. moongazer

it should be ln -1/2 =x and 3/2=y

51. jhonyy9

when by definition logb X so x need being allways positiv

52. hartnn

$$e^{i\pi}=-1$$ by euler's relation

53. jhonyy9

e^x - y = -2 e^x + y = 1 so and for these equations ?

54. hartnn

but if x is imaginary, y must be imaginary also, right ?

55. jhonyy9

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

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

I think my teacher doesn't need to know the decimal value of x.

58. hartnn

ok, then write that....

59. moongazer

ok, thanks for the answers :)

60. moongazer

why is it ln (negative number) is undefined?

61. jhonyy9

by domain definition of logarithms

62. jhonyy9

you have learned it sure --- i think ,yes ?

63. moongazer

ohh yes

64. moongazer

because of the graph

65. jhonyy9

good so than all is understood it right good luck bye

66. moongazer

Thanks :) do you know how to decompose a function?

67. jhonyy9

how do you think it ?

68. moongazer

partial fractions the one with cases. I got an odd looking expression

69. moongazer

i'll post it