## aroub 3 years ago Simplify using the law of exponents:

1. aroub

$\frac{ \sqrt[6]{c^5d} }{ \sqrt[4]{c^3d^3} }$

2. anonymous

$c^{m/n}=\sqrt[n]{c^m}$

3. aroub

- I never actually solved a question using " the laws of exponents" I just convert the final answer.. So, if you could just show me how or the steps of solving it please :)

4. anonymous

$\frac{ c^{5/6}d^{1/6} }{ c^{3/4}d^{3/4} }$ so we convert any radical to exponent using the rule above

5. ParthKohli

The only law that'd help nice would be:$\large \rm \sqrt[n]{x^m} = x^{m \over n}$

6. aroub

I know that rule! But then what?

7. ParthKohli

Also, you gotta use:$\rm {a^x \over a^y} = {a^{x - y}}$and$\rm {a^x}\cdot {a^y} = a^{x + y}$

8. ParthKohli

9. anonymous

$(\frac{ c^{5/6} }{ c^{3/4} })(\frac{ d^{1/6} }{ d^{3/4} })$

10. anonymous

can you use the rule by @ParthKohli $\frac{ a^m }{ a^n }=a^{m-n}$

11. aroub

Jonask, just continue solving please :) All I need is the steps!

12. anonymous

$c^{5/6-3/4}d^{1/6-3/4}$

13. anonymous

same base=subtract powers

14. aroub

cant you just add the numerator with numerator and the denominator with the denominator?

15. anonymous

can you illustrate

16. anonymous

no we can only addif they are both having same base

17. aroub

cd^5/6+1/6

18. aroub

they have the same base

19. anonymous

no c and d are not the same variable $\frac{ c^n*c^m }{ d^nd^m }=\frac{ c^{m+n} }{ d^{m+n} }$

20. aroub

Ahh, that's true! Sorry :]

21. anonymous

22. anonymous

$c^{-2/24}d^{-14/12}$ is the answer lets try this one$\frac{ x^3y^{-2} }{ \sqrt[3]{x^2y^2}}$

23. anonymous

@aroub sorry for this question ,,so why is this like difficile for you ?

24. aroub

No, it's okay! I'll tell you why, right there's two ways of solving this? First you can either solve it by keeping the radicals and then the final answer you can convert it OR by using the laws of exponents.. I always solve it using the first one -I find it easier. But then I found it that I have to solve it using the laws of exponents, so here I am >.<

25. anonymous

,,law of exponents" what method of solving is this ?

26. aroub

You convert EVERYTHING to exponents!

27. anonymous

yes than this mean very very easy

28. anonymous

begin with sqrtx =x^(1/2)

29. aroub

Lol, no to me -.- Well, I know how to convert them to exponents but then I have nooo ideaa! Anyway, thanks everyoonee! I gtg :D

30. anonymous

or a^(3/2) ------- = a^((3/2)-(1/3)) a^(1/3) right ?

31. anonymous

so and i think this will be the law of exponents method sure yes ?

32. aroub

Yup!

33. anonymous

ok good luck bye

34. aroub

Thanks! Bye =D

35. phi

I assume you have the answer, but to finish this post: $C^{(\frac{5}{6}-\frac{3}{4})}\cdot D^{(\frac{1}{6}-\frac{3}{4})}$ C's exponent is $\frac{5}{6}-\frac{3}{4} =\frac{5}{6} \cdot \frac{2}{2}-\frac{3}{4} \cdot \frac{3}{3}= \frac{10}{12}- \frac{9}{12}= \frac{1}{12}$ D's exponent is $\frac{1}{6}-\frac{3}{4} =\frac{1}{6} \cdot \frac{2}{2}-\frac{3}{4} \cdot \frac{3}{3}= \frac{2}{12}- \frac{9}{12}= -\frac{7}{12}$ you now hve $C^{\frac{1}{12}}D^{-\frac{7}{12}}= (\frac{C}{D^7})^\frac{1}{12}$ or $\sqrt[12]{\frac{C}{D^7}}$

36. aroub

That's the answer that I actually wanted! @phi thanks a looooottt ^_^