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anonymous

  • 5 years ago

how do i solve these kinds of problems? x 3/2 = 64

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  1. anonymous
    • 5 years ago
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    on my paper it looks like x with exponent 3 and exponent 2 = 64.... not sure how to put that on here

  2. asnaseer
    • 5 years ago
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    \[x^{\frac{3}{2}}=64\]is that right?

  3. anonymous
    • 5 years ago
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    yes

  4. asnaseer
    • 5 years ago
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    ok, firstly note that \(x^{\frac{1}{2}}\) means square root of x

  5. anonymous
    • 5 years ago
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    ok

  6. asnaseer
    • 5 years ago
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    so if we square both sides we get:\[(x^{\frac{3}{2}})^2=64^2\]therefore:\[x^3=64^2\]

  7. asnaseer
    • 5 years ago
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    now:\[64=2^6\]

  8. asnaseer
    • 5 years ago
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    so we can write this as:\[x^3=64^2=(2^6)^2=2^{12}\]

  9. anonymous
    • 5 years ago
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    wait i got confused, let me look at what you put

  10. asnaseer
    • 5 years ago
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    ok

  11. asnaseer
    • 5 years ago
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    I am using the rule:\[(x^a)^b=x^{ab}\]

  12. anonymous
    • 5 years ago
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    why is 64 squared 2 to 6th and not just 8?

  13. asnaseer
    • 5 years ago
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    \[64=2^6\]therefore:\[64^2=(2^6)^2=2^{12}\]

  14. asnaseer
    • 5 years ago
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    do you understand?

  15. anonymous
    • 5 years ago
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    no. 'im still trying to understand the squared part

  16. asnaseer
    • 5 years ago
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    ok, you agree:\[64=2^6\]

  17. anonymous
    • 5 years ago
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    ok

  18. asnaseer
    • 5 years ago
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    so we can substitute \(2^6\) where ever we see 64. so we can write:\[64^2=(64)^2=(2^6)^2=2^{12}\]

  19. asnaseer
    • 5 years ago
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    making sense?

  20. anonymous
    • 5 years ago
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    sort of

  21. anonymous
    • 5 years ago
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    i mean i understand the substituting part but i still dont understand the 64 squared = 2 to the 12th

  22. asnaseer
    • 5 years ago
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    are you familiar with the rule:\[(x^a)^b=x^{ab}\]

  23. anonymous
    • 5 years ago
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    yes

  24. anonymous
    • 5 years ago
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    but i just learned it, not a pro yet

  25. asnaseer
    • 5 years ago
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    :)

  26. asnaseer
    • 5 years ago
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    ok, lets try a different approach...

  27. anonymous
    • 5 years ago
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    ok, im sorry and thank you for being patient with me

  28. asnaseer
    • 5 years ago
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    \[64=2^6\]so we can write:\[64^p=(64)^p=(2^6)^p=2^{6p}\]

  29. asnaseer
    • 5 years ago
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    in our case p=2

  30. asnaseer
    • 5 years ago
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    we could also have written it as:\[64=8^2\]therefore:\[64^2=(8^2)^2=8^4\]

  31. asnaseer
    • 5 years ago
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    and then used:\[8=2^3\]therefore:\[64^2=8^4=(8)^4=(2^3)^4=2^{12}\]

  32. anonymous
    • 5 years ago
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    aha, and 8 to the 4th is the same as 2 to the 12th....ok got it now

  33. asnaseer
    • 5 years ago
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    yes - well done - I think the fog is beginning to clear :-)

  34. asnaseer
    • 5 years ago
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    ok, so lets review our last step...

  35. anonymous
    • 5 years ago
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    ok

  36. asnaseer
    • 5 years ago
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    we got to:\[x^3=64^2=(2^6)^2=2^{12}\]this should now be clear - yes?

  37. anonymous
    • 5 years ago
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    yes

  38. asnaseer
    • 5 years ago
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    ok, so next we take cube roots of both sides to get:\[(x^3)^{\frac{1}{3}}=(2^{12})^{\frac{1}{3}}\]therefore:\[x=2^4=16\]

  39. anonymous
    • 5 years ago
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    hold on..

  40. anonymous
    • 5 years ago
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    ok, i was doing the calculating to follow along with you

  41. asnaseer
    • 5 years ago
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    no problem - the main thing is to ensure you understand the concepts

  42. anonymous
    • 5 years ago
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    yes well you explained it great, thanks

  43. asnaseer
    • 5 years ago
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    you are very welcome. is it all clear now or did you want more explanation?

  44. anonymous
    • 5 years ago
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    i think i got it now, i was just stuck on the squared part. makes sense now. Thank you very much

  45. asnaseer
    • 5 years ago
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    ok - I'm glad I was enable to assist you in your mathematical quest :p - all the best...

  46. asnaseer
    • 5 years ago
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    *able (not enable)

  47. anonymous
    • 5 years ago
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    gotcha... thanks again :)

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