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anonymous

  • 5 years ago

solve :- log(cubic root of 3X-1) + log(cubic root of 3x+1) =log 20-1

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  1. anonymous
    • 5 years ago
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    \[\log \sqrt[3]{3X-1} + \log \sqrt[3]{3X+1} =\log 20-1\]

  2. amistre64
    • 5 years ago
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    emun can do it :)

  3. anonymous
    • 5 years ago
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    yeah

  4. anonymous
    • 5 years ago
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    give me a minute to finish it

  5. anonymous
    • 5 years ago
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    fast plz

  6. anonymous
    • 5 years ago
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  7. amistre64
    • 5 years ago
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    i mighta missed it lo

  8. anonymous
    • 5 years ago
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    not sure. That is what i got. Correct me if im wrong.

  9. anonymous
    • 5 years ago
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    wrong

  10. amistre64
    • 5 years ago
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    1.27614 maybe lol

  11. anonymous
    • 5 years ago
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    no

  12. anonymous
    • 5 years ago
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    1 , -1

  13. anonymous
    • 5 years ago
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    you have to isolate x now

  14. anonymous
    • 5 years ago
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    why U made the power 1/3 to 3

  15. anonymous
    • 5 years ago
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    answer me

  16. anonymous
    • 5 years ago
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    ahhh ok. Yeah definitively im wrong

  17. anonymous
    • 5 years ago
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    gonna try to solve it.

  18. anonymous
    • 5 years ago
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    U R welcome

  19. anonymous
    • 5 years ago
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    Sorry i modified that, but i didnt get 1, -1

  20. anonymous
    • 5 years ago
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    so what did U get

  21. anonymous
    • 5 years ago
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    hah

  22. anonymous
    • 5 years ago
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    answer me

  23. anonymous
    • 5 years ago
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    something that you would make fun of. I have to have diner now. I'll answer you later. just be pacient

  24. anonymous
    • 5 years ago
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    Here you have

  25. anonymous
    • 5 years ago
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    wrong

  26. anonymous
    • 5 years ago
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    look

  27. anonymous
    • 5 years ago
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    Can you realize in what part I mess up?

  28. anonymous
    • 5 years ago
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    log 20/10 = log 2

  29. anonymous
    • 5 years ago
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    I didt do that anywhere

  30. anonymous
    • 5 years ago
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    I dont need that process

  31. anonymous
    • 5 years ago
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    sorry at the end

  32. anonymous
    • 5 years ago
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    \[(3X-1)(3X+1) = 8\]

  33. anonymous
    • 5 years ago
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    it must be

  34. anonymous
    • 5 years ago
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    You are right, but the mistake is at the begining. Gonna try to solve it again

  35. anonymous
    • 5 years ago
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    \[9 x ^{2 } + 1 = 8\]

  36. anonymous
    • 5 years ago
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    \[9 X ^{2} = 9\]

  37. anonymous
    • 5 years ago
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    X = 1 , -1 s.s = {1}

  38. anonymous
    • 5 years ago
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    -1 ref..

  39. anonymous
    • 5 years ago
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  40. anonymous
    • 5 years ago
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    yeah you were right :)

  41. anonymous
    • 5 years ago
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    Hope I was helpful here

  42. anonymous
    • 5 years ago
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    U r always

  43. anonymous
    • 5 years ago
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    but

  44. anonymous
    • 5 years ago
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    but ... too slow

  45. anonymous
    • 5 years ago
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    noooooooooooooooo

  46. anonymous
    • 5 years ago
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    \[10^{\log20} *10^{-1}\]

  47. anonymous
    • 5 years ago
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    or \[10^{\log20} - 10^{-1}\]

  48. anonymous
    • 5 years ago
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    the first one, because \[a ^{m}a ^{n}=a ^{m+n}\]

  49. anonymous
    • 5 years ago
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    clear?

  50. anonymous
    • 5 years ago
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    ok

  51. anonymous
    • 5 years ago
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    can you give me a medal please :) I think I deserve it

  52. anonymous
    • 5 years ago
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    but how

  53. myininaya
    • 5 years ago
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  54. anonymous
    • 5 years ago
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    ^ *cough*

  55. anonymous
    • 5 years ago
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    forget something myininaya?

  56. myininaya
    • 5 years ago
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    lol the 1 (x-1)(x+1)=0

  57. anonymous
    • 5 years ago
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    no, you are saying the solution is x=+-1 , are you sure :P

  58. myininaya
    • 5 years ago
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    well its not -1 because we can't have log of negative number

  59. myininaya
    • 5 years ago
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    so its just x=1 i was hoping he would know to do that but you are right i should had said that

  60. anonymous
    • 5 years ago
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    but I know

  61. anonymous
    • 5 years ago
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    this all in my studies

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