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anonymous

  • 5 years ago

(x^1/6y^1/3)^-18

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  1. anonymous
    • 5 years ago
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    so I will do numerator and denominator separately

  2. anonymous
    • 5 years ago
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    okay thank you :0]

  3. anonymous
    • 5 years ago
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    numerator --(x^1)^-18=x^-18

  4. anonymous
    • 5 years ago
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    now what is this numerator go to?

  5. anonymous
    • 5 years ago
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    denominator -- (6y^1/3)^-18 =(6^-18)*(y^-6)

  6. anonymous
    • 5 years ago
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    put them together (x^-18)/(6^-18)(y^-6)

  7. anonymous
    • 5 years ago
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    okay break this down a bit. for the numerator where did you get the x^1?

  8. anonymous
    • 5 years ago
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    wasn't that in the question?

  9. anonymous
    • 5 years ago
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    no its x^1/6

  10. anonymous
    • 5 years ago
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    no its x^1/6

  11. anonymous
    • 5 years ago
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    oh, I misread the question, sorry

  12. anonymous
    • 5 years ago
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    its okay it was just hiding is all. haha

  13. anonymous
    • 5 years ago
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    (x^(1/6) * y^(1/3))^-18 x^-3 y^-6

  14. anonymous
    • 5 years ago
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    I just multiplied the exponents

  15. anonymous
    • 5 years ago
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    theres no multiplication sign between x and y

  16. anonymous
    • 5 years ago
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    when there is nothing, it is assumed that they are being multiplied

  17. anonymous
    • 5 years ago
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    oh okay. so for the exponents do you just make the 3 qnd 6 equal to -18?

  18. anonymous
    • 5 years ago
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    When you have one exponent on another like this \[(a^x)^y\] You would multiply the exponents x*y

  19. anonymous
    • 5 years ago
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    can be rewritten as a^(x*y)

  20. anonymous
    • 5 years ago
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    xy^1/18?

  21. anonymous
    • 5 years ago
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    so in your case (x^(1/6) * y^(1/3))^-18

  22. anonymous
    • 5 years ago
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    nvm i got it x^-3 and y^-6 right?

  23. anonymous
    • 5 years ago
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    you are right

  24. anonymous
    • 5 years ago
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    is that as simpilified as im going to get?

  25. anonymous
    • 5 years ago
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    Yes

  26. anonymous
    • 5 years ago
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    alrighty thank you for helping me :0]

  27. anonymous
    • 5 years ago
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    btw do you know that negative exponent could be written as positive if you put in denominator

  28. anonymous
    • 5 years ago
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    no i didn't but that might be important right?

  29. anonymous
    • 5 years ago
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    yes say you have something like x^-a you can rewrite it as 1/x^a

  30. anonymous
    • 5 years ago
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    1/ over the equation

  31. anonymous
    • 5 years ago
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    okay yeah i saw that in a video i was recently watching okay thanks a lot!! that will help towards my test. i really really apperciate everything. :0]

  32. anonymous
    • 5 years ago
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    also remember this x^a * y^b=z^(a+b) x^a/y^b= z^(a-b)

  33. anonymous
    • 5 years ago
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    special cases??

  34. anonymous
    • 5 years ago
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    no it is always the case Trying solving this x^2 * y^4

  35. anonymous
    • 5 years ago
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    z^6?

  36. anonymous
    • 5 years ago
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    Yes

  37. anonymous
    • 5 years ago
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    I forgot to say this but x and y must be same

  38. anonymous
    • 5 years ago
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    sorry my computer is running slow. in the cases you just showed me is that what your talking about?

  39. anonymous
    • 5 years ago
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    yes

  40. anonymous
    • 5 years ago
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    yes

  41. anonymous
    • 5 years ago
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    2^3 * 2^2 = 8 * 4= 32 or 2^(3+2) = 2^5= 32

  42. anonymous
    • 5 years ago
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    Will you test have log

  43. anonymous
    • 5 years ago
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    okay for a the problem im working on now (x^1/3 /y^-2/3)^9 do i divide first?

  44. anonymous
    • 5 years ago
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    log? like on the calculator?

  45. anonymous
    • 5 years ago
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    since you do not know if x and y are same don't devide anything

  46. anonymous
    • 5 years ago
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    Just do what we did on last problem

  47. anonymous
    • 5 years ago
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    i did i got x^3/y^-6 but my book says the answer os x^3y^6

  48. anonymous
    • 5 years ago
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    -2/3 * 9

  49. anonymous
    • 5 years ago
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    i got -6 and i did the same for 1/3 and got 3

  50. anonymous
    • 5 years ago
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    Both you and the book are right

  51. anonymous
    • 5 years ago
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    1/a^-x = a^x 1/a^x = a^-x

  52. anonymous
    • 5 years ago
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    hmmm? but i have the x^3 on top how can you fit a 1 by itself?

  53. anonymous
    • 5 years ago
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    This is what you got x^3/y^-6 using 1/a^-x = a^x 1/a^x = a^-x x^3 * 1/y^-6 x^3 * y^6

  54. anonymous
    • 5 years ago
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    oh okay that makes more sense.

  55. anonymous
    • 5 years ago
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    1 * x^3 just x^3

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