## anonymous 5 years ago (-4x^2y)^3 I really need help, I still don't understand.

1. anonymous

$(-4x^{2y})^{3} is this what you mean?$

2. anonymous

$(ab^{c})^{d} = a^{d} b ^{cd}$

3. anonymous

so it will be $-4^{3}x ^{6y}$

4. anonymous

= -64x^6y

5. anonymous

remember that y has an implicit exponent of 1, and 1*3 = 3.

6. anonymous

no i meant (-4x to the second power than y)^3 the y is not part of the exponent

7. anonymous

oh, so it's $(-4x ^{2}y)^{?}$

8. anonymous

the concept is the same, -4^3 x^6 y^3

9. anonymous

or -64x^6 y^3

10. anonymous

I think the question is :$(-4x^2y)^3$ right?

11. anonymous

yes thats right, i just dont understand how to solve it

12. anonymous

As I said, when raising a product to a power, you multiply the exponents of the factors. So the 3 would be multiplied to the power of the -4 factor, and the power of the x factor, and the power of the y factor.

13. anonymous

alright, first of all you've got to mulitply the power 3 with the given between the brackets and you'll get ^_^: $=(-4)^3 x^6 y^3 = 48x^6y^3$ is it clearer now ? :)

14. anonymous

$(-4)^3 \ne 48$ $(-4)^3 = -64$

15. anonymous

lol! sry >_< silly mistake , thank you polpak :)

16. anonymous

do you understand it now courtney ^_^?

17. anonymous

just substitute the power to each :)

18. anonymous

just pass the power out to each*

19. anonymous

yes, kind of but there is another problem now that is in the same format but is a fraction and has that equal sign with a / through it. grr i hate math lol

20. anonymous

don't say that, look at the problem , write it down on paper and try to use tha same concept like the one here :) The problem never gets harder, it just gets more detailed ^_^

21. anonymous

nothing changes, just apply the same rule :)

22. anonymous

The rule applies equally to division. The only part that gets messy is when you have a sum that you're raising to a power.

23. anonymous

im trying to stay positive, thank you so much for your help you two are life savers! im writing all the work you just sent as my work shown but what part is the part that i put in the space for the answer to the problem?

24. anonymous

Try to see if you can figure it out from what we've said. We'll check your answer.

25. anonymous

np :)

26. anonymous

ok im working it out now than ill post it on here, thank you again.

27. anonymous

$(((3p ^{3}v ^{4} \ s ^{4}))^{2}$

28. anonymous

ok this is a fraction the s^4 is on the bottom and it also says s with the equal sign / 0

29. anonymous

i got 9p^6y^8 over s^8

30. anonymous

So.. $(3p^3v^4/s^4)^2, s \ne 0$

31. anonymous

Then yes, that is the correct answer. Good job!

32. anonymous

yayyy thank you, even my professor cant teach me math and over the computer you got right into my head lol

33. anonymous

It may be somewhat helpful to expand the exponents for a while to see what is happening. $(3p^3v^4/s^4)^2 = [(3pppvvvv)/(ssss)]^2$ $= [(3pppvvvv)(3pppvvvv)]/[(ssss)(ssss)]$ $= [(3*3)ppppppvvvvvvvv]/[ssssssss]$ $= 9p^6v^8/s^8$

34. anonymous

that acutally helps alot, can i ask you on more question?

35. anonymous

Of course!

36. anonymous

ok so i know how to do this in a normal way but they want it in exponential form and im confused. the first is (5^3)^3 and the next is in fraction form which is even worse lol (r/s)^6 (s$\neq$0)

37. anonymous

So try to expand it out. $(5^3)^3 = (5^3)(5^3)(5^3)$ Now how would you expand that further?

38. anonymous

five times five times five than add the exponents?

39. anonymous

$5^3 = 5*5*5$ $\rightarrow 5^3*5^3*5^3 = (5*5*5)*(5*5*5)*(5*5*5)$ $=5^?$

40. anonymous

5^9?

41. anonymous

Yep.

42. anonymous

awesome now to the fraction, can you help me with that one?

43. anonymous

Now we can see that the faster way to arrive at the same result would be to simply : $(5^3)^3 = 5^{3*3} = 5^9$

44. anonymous

okay so you multiply the exponents too?

45. anonymous

Well, remember when we expanded it out, we said that we had three groups with three 5's each. So how many 5's do we have multiplied together? 3*3 right?

46. anonymous

right, im trying to keep up im just so lost

47. anonymous

So if you're not comfortable with multiplying the exponents, just try expanding the next one out for yourself. $a^6 = a*a*a*a*a*a$ $\rightarrow (r/s)^6 =\ ?$

48. anonymous

We'll just go one step at a time.

49. anonymous

would i seperate them and do r*r*r*r*r*r and s*s*s*s*s*s?

50. anonymous

No, just multiply the whole thing.

51. anonymous

so r/s*r/s*r/s*r/s*r/s*r/s

52. anonymous

Yes, but keep the parens so it's more readable. $(r/s)*(r/s)*(r/s)*(r/s)*(r/s)*(r/s)$ And when we multiply two fractions, we do what?

53. anonymous

top times top bottom times bottom?

54. anonymous

yep

55. anonymous

so its (r^6)/(s^6)

56. anonymous

Yes. And here again we can arrive at this answer in a more direct way once we understand the process $(r/s)^6 = (r^1/s^1)^6 = r^{1*6}/s^{1*6} = r^6/s^6$

57. anonymous

that would be the answer correct since we do not have the values for r and s? what about the =/ thing?

58. anonymous

They have to make the restriction $s \ne 0$ because if s is 0 then you're dividing by 0 which is not defined. That answer is correct regardless of what r and s is. If we know more about them we might be able to simplify further (if s was 2 and r was 4 for example). But even in that case our answer is correct. It just might not be in the form they're looking for.

59. anonymous

yayyy you are so amazing and kind hearted. that you so so much!

60. anonymous

You're very welcome.