anonymous
  • anonymous
(^3sqrtc^7d^4)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
so the problem is this? \[\LARGE \left(\sqrt[3]{c^7d^4}\right)^2\]
anonymous
  • anonymous
Yes
jim_thompson5910
  • jim_thompson5910
and they want you to simplify? or rewrite into rational exponent form?

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anonymous
  • anonymous
simplify
jim_thompson5910
  • jim_thompson5910
ok are you familiar with converting radical form to rational exponent form?
anonymous
  • anonymous
not at all
jim_thompson5910
  • jim_thompson5910
I'm going to use this rule \[\LARGE \sqrt[n]{x^m} = x^{m/n}\] hopefully it looks familiar
anonymous
  • anonymous
ok... walk me through how to solve this
jim_thompson5910
  • jim_thompson5910
so we use that rule to go from \[\LARGE \sqrt[3]{c^7d^4}\] to \[\LARGE \left(c^7d^4\right)^{1/3}\]
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
which is why \[\LARGE \left(\sqrt[3]{c^7d^4}\right)^2\] is the same as \[\LARGE \left(\left(c^7d^4\right)^{1/3}\right)^2\]
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
then we multiply the exponents \[\LARGE \left(\left(c^7d^4\right)^{1/3}\right)^2\] \[\LARGE \left(c^7d^4\right)^{1/3*2}\] \[\LARGE \left(c^7d^4\right)^{2/3}\]
anonymous
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
so do you see how I got \[\LARGE \left(c^7d^4\right)^{2/3}\]
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
now we multiply the inner exponents by the outer exponent 2/3 \[\LARGE \left(c^7d^4\right)^{2/3}\] \[\LARGE c^{7*2/3}d^{4*2/3}\] \[\LARGE c^{14/3}d^{8/3}\]
jim_thompson5910
  • jim_thompson5910
when you divide 14/3, what is the quotient and remainder?
anonymous
  • anonymous
I think it would be 14 is youre remainder and 3 is youre quotient
jim_thompson5910
  • jim_thompson5910
14/3 = 4 remainder 3 4 is the quotient, 3 is the remainder
jim_thompson5910
  • jim_thompson5910
that leads us to \[\LARGE c^{14/3} = c^4\sqrt[3]{c^2}\]
jim_thompson5910
  • jim_thompson5910
similarly, \[\LARGE d^{8/3} = d^2\sqrt[3]{d^2}\]
jim_thompson5910
  • jim_thompson5910
so overall \[\LARGE c^{14/3}d^{8/3} = c^4\sqrt[3]{c^2}*d^2\sqrt[3]{d^2}\] \[\LARGE c^{14/3}d^{8/3} = c^4d^2\sqrt[3]{c^2d^2}\]
jim_thompson5910
  • jim_thompson5910
sorry I meant to say "14/3 = 4 remainder 2" (not remainder 3)
anonymous
  • anonymous
Thank you :) could you help with a couple more
jim_thompson5910
  • jim_thompson5910
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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