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## TrojanPoem one year ago if 6^ (-1/a) = 3^(1/b) = 2^ (1/g)

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1. TrojanPoem

Prove that |dw:1434806977129:dw|

2. dan815

is that the determinant

3. TrojanPoem

Yeah , no expansion

4. ganeshie8

what do we know about a,b,g are they integers

5. TrojanPoem

6^ (-1/a) = 3^(1/b) = 2^ (1/g)

6. dan815

|dw:1434807323802:dw|

7. TrojanPoem

I thought of adding logs 6^ (-1/a) = 3^(1/b) = 2^ (1/g) and get out a/b , b/g , g/a however failed

8. dan815

ya i was about to say i think we need to get simpler expressions with logs

9. dan815

probably mess around with that more

10. TrojanPoem

|dw:1434807544755:dw|

11. TrojanPoem

|dw:1434807600411:dw|

12. TrojanPoem

|dw:1434807629601:dw|

13. dan815

oh how about doing trace stuff

14. dan815

or eigen value stuff, to show this is dependant

15. TrojanPoem

Show out your work !

16. dan815

i think we can get those expressions for a b g, by solving for eigen values given your matrix,

17. dan815

|dw:1434807849696:dw|

18. ganeshie8

how is finding eigenvalues any easier than row reduction followed by expansion

19. dan815

ok i guess not lol xD

20. dan815

that involves expansion right in there

21. TrojanPoem

what is eigenvalues ?

22. TrojanPoem

You can easily get its value without expanding here you go

23. ganeshie8

it is a fancy term whihch linear algebra freaks use a lot ;p

24. dan815

lool

25. TrojanPoem

:D

26. dan815

did u try t work with this expression

27. dan815

|dw:1434808112422:dw|

28. TrojanPoem

Yeah, here is the result -A/B = Log6/ Log 3

29. TrojanPoem

Log3 / Log 2 = B/G

30. TrojanPoem

Log6/ Lo 2 = -A/G

31. TrojanPoem

I even added them to the det

32. TrojanPoem

Any ideas ?

33. dan815

forget the divisions for now, how about the additions can u get simp exp for that

34. dan815

|dw:1434808467873:dw|

35. TrojanPoem

You don't want them divided ? maybe you can right log3/log 6 as log3_6

36. dan815

|dw:1434808505436:dw|

37. TrojanPoem

Hmm C/in2 + D/in3 how did you separated it into two ? C/ in2 + in3 + D/in2 + in3

38. dan815

i didnt do it, but we can do it, just look at the possibilities, that is partial fractions

39. dan815

a/(e+f) = c/e + d/f there is always a way to find htis

40. TrojanPoem

Got it

41. dan815

where c and d are in terms of a

42. TrojanPoem

we have to find the constants c, d right ?

43. dan815

yea

44. dan815

ii dont see where this goes though -.-

45. ganeshie8

It is easy to show that $$a+b+g=0$$

46. ganeshie8

|dw:1434808938502:dw|

47. TrojanPoem

B = GLog3 / Log2 A = -GLo6/Log2 A + b +g = Glo3/log2 - GLog6/Log2 + g = GLog3 - Glo3 - Glo2 + Glo2 / Log2 = 0/Log2 = 0

48. TrojanPoem

right a+ b + g = 0

49. ganeshie8

Looks good!

50. TrojanPoem

Now ideas ?

51. dan815

well its solved then

52. ganeshie8

Same thing but below looks less messier $$6^{-1/a} = 3^{1/b} = 2^{1/g} = k$$ $$a = \frac{-\ln 6}{k}$$ $$b = \frac{\ln 3}{k}$$ $$g = \frac{\ln 2}{k}$$ $$a+b+g = 0$$

53. dan815

a b and g are not = 0 so

54. dan815

we can safely factor out a+b+g out off everything

55. dan815

nice

56. TrojanPoem

Now ideas ?

57. ganeshie8

looks we're done! aren't we ?

58. dan815

|dw:1434809493246:dw|

59. TrojanPoem

LOL, I understood

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