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TrojanPoem

  • one year ago

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

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  1. TrojanPoem
    • one year ago
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    Prove that |dw:1434806977129:dw|

  2. dan815
    • one year ago
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    is that the determinant

  3. TrojanPoem
    • one year ago
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    Yeah , no expansion

  4. ganeshie8
    • one year ago
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    what do we know about a,b,g are they integers

  5. TrojanPoem
    • one year ago
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    6^ (-1/a) = 3^(1/b) = 2^ (1/g)

  6. dan815
    • one year ago
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    |dw:1434807323802:dw|

  7. TrojanPoem
    • one year ago
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    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
    • one year ago
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    ya i was about to say i think we need to get simpler expressions with logs

  9. dan815
    • one year ago
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    probably mess around with that more

  10. TrojanPoem
    • one year ago
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    |dw:1434807544755:dw|

  11. TrojanPoem
    • one year ago
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    |dw:1434807600411:dw|

  12. TrojanPoem
    • one year ago
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    |dw:1434807629601:dw|

  13. dan815
    • one year ago
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    oh how about doing trace stuff

  14. dan815
    • one year ago
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    or eigen value stuff, to show this is dependant

  15. TrojanPoem
    • one year ago
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    Show out your work !

  16. dan815
    • one year ago
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    i think we can get those expressions for a b g, by solving for eigen values given your matrix,

  17. dan815
    • one year ago
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    |dw:1434807849696:dw|

  18. ganeshie8
    • one year ago
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    how is finding eigenvalues any easier than row reduction followed by expansion

  19. dan815
    • one year ago
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    ok i guess not lol xD

  20. dan815
    • one year ago
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    that involves expansion right in there

  21. TrojanPoem
    • one year ago
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    what is eigenvalues ?

  22. TrojanPoem
    • one year ago
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    You can easily get its value without expanding here you go

  23. ganeshie8
    • one year ago
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    it is a fancy term whihch linear algebra freaks use a lot ;p

  24. dan815
    • one year ago
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    lool

  25. TrojanPoem
    • one year ago
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    :D

  26. dan815
    • one year ago
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    did u try t work with this expression

  27. dan815
    • one year ago
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    |dw:1434808112422:dw|

  28. TrojanPoem
    • one year ago
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    Yeah, here is the result -A/B = Log6/ Log 3

  29. TrojanPoem
    • one year ago
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    Log3 / Log 2 = B/G

  30. TrojanPoem
    • one year ago
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    Log6/ Lo 2 = -A/G

  31. TrojanPoem
    • one year ago
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    I even added them to the det

  32. TrojanPoem
    • one year ago
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    Any ideas ?

  33. dan815
    • one year ago
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    forget the divisions for now, how about the additions can u get simp exp for that

  34. dan815
    • one year ago
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    |dw:1434808467873:dw|

  35. TrojanPoem
    • one year ago
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    You don't want them divided ? maybe you can right log3/log 6 as log3_6

  36. dan815
    • one year ago
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    |dw:1434808505436:dw|

  37. TrojanPoem
    • one year ago
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    Hmm C/in2 + D/in3 how did you separated it into two ? C/ in2 + in3 + D/in2 + in3

  38. dan815
    • one year ago
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    i didnt do it, but we can do it, just look at the possibilities, that is partial fractions

  39. dan815
    • one year ago
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    a/(e+f) = c/e + d/f there is always a way to find htis

  40. TrojanPoem
    • one year ago
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    Got it

  41. dan815
    • one year ago
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    where c and d are in terms of a

  42. TrojanPoem
    • one year ago
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    we have to find the constants c, d right ?

  43. dan815
    • one year ago
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    yea

  44. dan815
    • one year ago
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    ii dont see where this goes though -.-

  45. ganeshie8
    • one year ago
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    It is easy to show that \(a+b+g=0\)

  46. ganeshie8
    • one year ago
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    |dw:1434808938502:dw|

  47. TrojanPoem
    • one year ago
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    B = GLog3 / Log2 A = -GLo6/Log2 A + b +g = Glo3/log2 - GLog6/Log2 + g = GLog3 - Glo3 - Glo2 + Glo2 / Log2 = 0/Log2 = 0

  48. TrojanPoem
    • one year ago
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    right a+ b + g = 0

  49. ganeshie8
    • one year ago
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    Looks good!

  50. TrojanPoem
    • one year ago
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    Now ideas ?

  51. dan815
    • one year ago
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    well its solved then

  52. ganeshie8
    • one year ago
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    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
    • one year ago
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    a b and g are not = 0 so

  54. dan815
    • one year ago
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    we can safely factor out a+b+g out off everything

  55. dan815
    • one year ago
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    nice

  56. TrojanPoem
    • one year ago
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    Now ideas ?

  57. ganeshie8
    • one year ago
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    looks we're done! aren't we ?

  58. dan815
    • one year ago
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    |dw:1434809493246:dw|

  59. TrojanPoem
    • one year ago
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    LOL, I understood

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