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

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

Mathematics
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so the problem is this? \[\LARGE \left(\sqrt[3]{c^7d^4}\right)^2\]
Yes
and they want you to simplify? or rewrite into rational exponent form?

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Other answers:

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