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anonymous

  • one year ago

(^3sqrtc^7d^4)^2

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  1. jim_thompson5910
    • one year ago
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    so the problem is this? \[\LARGE \left(\sqrt[3]{c^7d^4}\right)^2\]

  2. anonymous
    • one year ago
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    Yes

  3. jim_thompson5910
    • one year ago
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    and they want you to simplify? or rewrite into rational exponent form?

  4. anonymous
    • one year ago
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    simplify

  5. jim_thompson5910
    • one year ago
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    ok are you familiar with converting radical form to rational exponent form?

  6. anonymous
    • one year ago
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    not at all

  7. jim_thompson5910
    • one year ago
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    I'm going to use this rule \[\LARGE \sqrt[n]{x^m} = x^{m/n}\] hopefully it looks familiar

  8. anonymous
    • one year ago
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    ok... walk me through how to solve this

  9. jim_thompson5910
    • one year ago
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    so we use that rule to go from \[\LARGE \sqrt[3]{c^7d^4}\] to \[\LARGE \left(c^7d^4\right)^{1/3}\]

  10. anonymous
    • one year ago
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    ok

  11. jim_thompson5910
    • one year ago
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    which is why \[\LARGE \left(\sqrt[3]{c^7d^4}\right)^2\] is the same as \[\LARGE \left(\left(c^7d^4\right)^{1/3}\right)^2\]

  12. anonymous
    • one year ago
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    ok

  13. jim_thompson5910
    • one year ago
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    then we multiply the exponents \[\LARGE \left(\left(c^7d^4\right)^{1/3}\right)^2\] \[\LARGE \left(c^7d^4\right)^{1/3*2}\] \[\LARGE \left(c^7d^4\right)^{2/3}\]

  14. anonymous
    • one year ago
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    ok

  15. jim_thompson5910
    • one year ago
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    so do you see how I got \[\LARGE \left(c^7d^4\right)^{2/3}\]

  16. anonymous
    • one year ago
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    yes

  17. jim_thompson5910
    • one year ago
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    now we multiply the inner exponents by the outer exponent 2/3 \[\LARGE \left(c^7d^4\right)^{2/3}\] \[\LARGE c^{7*2/3}d^{4*2/3}\] \[\LARGE c^{14/3}d^{8/3}\]

  18. jim_thompson5910
    • one year ago
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    when you divide 14/3, what is the quotient and remainder?

  19. anonymous
    • one year ago
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    I think it would be 14 is youre remainder and 3 is youre quotient

  20. jim_thompson5910
    • one year ago
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    14/3 = 4 remainder 3 4 is the quotient, 3 is the remainder

  21. jim_thompson5910
    • one year ago
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    that leads us to \[\LARGE c^{14/3} = c^4\sqrt[3]{c^2}\]

  22. jim_thompson5910
    • one year ago
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    similarly, \[\LARGE d^{8/3} = d^2\sqrt[3]{d^2}\]

  23. jim_thompson5910
    • one year ago
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    so overall \[\LARGE c^{14/3}d^{8/3} = c^4\sqrt[3]{c^2}*d^2\sqrt[3]{d^2}\] \[\LARGE c^{14/3}d^{8/3} = c^4d^2\sqrt[3]{c^2d^2}\]

  24. jim_thompson5910
    • one year ago
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    sorry I meant to say "14/3 = 4 remainder 2" (not remainder 3)

  25. anonymous
    • one year ago
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    Thank you :) could you help with a couple more

  26. jim_thompson5910
    • one year ago
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    I'll help with one more. Please post where it says "ask a question" so you can start a new post

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