## anonymous one year ago (^3sqrtc^7d^4)^2

1. jim_thompson5910

so the problem is this? $\LARGE \left(\sqrt[3]{c^7d^4}\right)^2$

2. anonymous

Yes

3. jim_thompson5910

and they want you to simplify? or rewrite into rational exponent form?

4. anonymous

simplify

5. jim_thompson5910

ok are you familiar with converting radical form to rational exponent form?

6. anonymous

not at all

7. jim_thompson5910

I'm going to use this rule $\LARGE \sqrt[n]{x^m} = x^{m/n}$ hopefully it looks familiar

8. anonymous

ok... walk me through how to solve this

9. jim_thompson5910

so we use that rule to go from $\LARGE \sqrt[3]{c^7d^4}$ to $\LARGE \left(c^7d^4\right)^{1/3}$

10. anonymous

ok

11. jim_thompson5910

which is why $\LARGE \left(\sqrt[3]{c^7d^4}\right)^2$ is the same as $\LARGE \left(\left(c^7d^4\right)^{1/3}\right)^2$

12. anonymous

ok

13. jim_thompson5910

then we multiply the exponents $\LARGE \left(\left(c^7d^4\right)^{1/3}\right)^2$ $\LARGE \left(c^7d^4\right)^{1/3*2}$ $\LARGE \left(c^7d^4\right)^{2/3}$

14. anonymous

ok

15. jim_thompson5910

so do you see how I got $\LARGE \left(c^7d^4\right)^{2/3}$

16. anonymous

yes

17. jim_thompson5910

now we multiply the inner exponents by the outer exponent 2/3 $\LARGE \left(c^7d^4\right)^{2/3}$ $\LARGE c^{7*2/3}d^{4*2/3}$ $\LARGE c^{14/3}d^{8/3}$

18. jim_thompson5910

when you divide 14/3, what is the quotient and remainder?

19. anonymous

I think it would be 14 is youre remainder and 3 is youre quotient

20. jim_thompson5910

14/3 = 4 remainder 3 4 is the quotient, 3 is the remainder

21. jim_thompson5910

that leads us to $\LARGE c^{14/3} = c^4\sqrt[3]{c^2}$

22. jim_thompson5910

similarly, $\LARGE d^{8/3} = d^2\sqrt[3]{d^2}$

23. jim_thompson5910

so overall $\LARGE c^{14/3}d^{8/3} = c^4\sqrt[3]{c^2}*d^2\sqrt[3]{d^2}$ $\LARGE c^{14/3}d^{8/3} = c^4d^2\sqrt[3]{c^2d^2}$

24. jim_thompson5910

sorry I meant to say "14/3 = 4 remainder 2" (not remainder 3)

25. anonymous

Thank you :) could you help with a couple more

26. jim_thompson5910

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