## anonymous 5 years ago Okay, lets try this again... [(3^-1)(x)(y^-4)(z^5)/(7x^-4)(y^2)(z)]^-2

1. anonymous

Haha this looks awfully familiar. Did you have a problem with the other answer?

2. anonymous

I got this: $[(x^5z^7)/(21y^6)]^-2$ so far

3. anonymous

Yes I did, but now I got stuck

4. anonymous

Mine was slightly different, the z value ended up being different from what you gave me, and I don't understand why.

5. anonymous

Z doesn't move. You just cancel it normally. 5-1 is 4. I didn't even pay attention to z until the very end.

6. anonymous

oh yea, thats right

7. anonymous

3^-1 right? so, where is 21 come from? I got: (x5z7)/(3y6)

8. anonymous

Inik the 3 moves to the bottom with the z, because it has the negative power. Then you multiply them to get 21.

9. anonymous

ooo... I see 7 as well. agree with 21

10. anonymous

Alright, l2theM, do you understand/have any more questions?

11. anonymous

Well, I still haven't simplified all of the way. Now that I got the above answer, how do I handle the ^-2? Do I move the top portion to the bottom, and the bottom portion to the top?

12. anonymous

*I did fix the z

13. anonymous

sorry it's messy

14. anonymous

Rain, how did you get the 3 on the bottom and 39 on the top?

15. anonymous

eh, 49 rather

16. anonymous

Yes. Everything gets flipped because they all become negative. Then you multiply all exponents by 2 and square the 21. You should get: $x ^{10} z ^{8} \div 441 y ^{12}$

17. anonymous

Rain, one that makes it incredibly complicated and confusing... And you also forgot to square the 3.

18. anonymous

I got 441y^12z^2/x^10z^10

19. anonymous

Yeah sorry I typed it backwards. Yours is right.

20. anonymous

I don't understand what you are doing with the z though. The z^5 should go to the bottom, and the z should go to the top. Why is it z^8 then?

21. anonymous

22. anonymous

and also, since I have a z^2 on the top, and a z^10 on the bottom, do I simplify them to z^1/z^5?

23. anonymous

Because you still cancel them. 5-1 is still 4 lol Once it is cancelled you square it, and since it is an exponent you multiply them, so 4x2=8.

24. anonymous

Rain, I don't think that is right.

25. anonymous

oh it's the whole thing to the power of -2. I only did the bottom. My bad, sorry!

26. anonymous

You can do it either way. If you multiply first, you get the z^2/z^10 and you cancel by doing 10-2 which is still 8 lol.

27. anonymous

Hahaha no worries. Just remember it applies to the whole thing. I like doing things the easy way, so thats why I did it in that order.

28. anonymous

but why are you subtracting them if they are on opposite sides of the equation? are you changing the z^2 to z^-2 and subtracting it from the larger z^10?

29. anonymous

Here, let me explain another way. How would you simplify $x ^{2} \div x$ ?

30. anonymous

Okay, what I did was I simplified everything in the equation and made it so all of the exponents were positive. Then I switch their locations so the ^-2 would become ^2, then I squared everything and got this: 441y^12z^2/x^10z^10

31. anonymous

That would be just x

32. anonymous

and I am sorry for the trouble, I just want to make sure I am doing this right. I have a final coming up, and I need to make a 95 or better, so really want to fully understand this.

33. anonymous

No worries at all. That's that this site is for! Well, you have that down, so, explain to me how you got that. And how would you do x^87/x?

34. anonymous

it would be x^87 with no denominator, right? I think I did the first sub-question wrong

35. anonymous

No you did the first one right. But what did you DO? How did you know that it was just x?

36. anonymous

And no, it would be x^86. so what did I do there? How did I get that answer?

37. anonymous

You must have just subtracted them. But I am thinking of an equivalent equation with the same bases like this: 4/2. 4 is the same as 2^2. So the equivalent equation would be 2^2/2. The answer would be 2 no matter what.

38. anonymous

Oh, you DID subtract the base from the numerator. x*x/x would mean 1 x goes away, which would be subtracting, right?

39. anonymous

Quantum, I got your answer but flipped. $411y ^{12} \div x ^{10} z ^{8}$What did I do wrong?

40. anonymous

I think you got it right this time.

41. anonymous

Normally I get an answer sheet with my practice tests, but my final doesn't give me one so I have no idea what the true answer is.

42. anonymous

Hm it should be right. http://www.wolframalpha.com/input/?i=%5B%5B%283^-1%29%28x%29%28y^-4%29%28z^5%29%5D%2F%5B%287x^-4%29%28y^2%29%28z%29%5D%5D^-2 Copy&paste that link

43. anonymous

I don't know how to do it by that method Rain, that looks completely foreign to me.

44. anonymous

Yes, that's right. You subtract them! Does that make sense now?

45. anonymous

Yes, it makes sense when you have x^2/x, but when you have x/x^2, that would make the problem equal x too, right?

46. anonymous

x/x=x?

47. anonymous

You subtract the bigger number by the smaller number, and whichever one was bigger, that's where it goes. Like if the x^87 was on bottom and x was on top, it would be 1/x^86. it's the same thing.

48. anonymous

No, x/x is one. Because both cancel each other. It's like 4/4 or 7/7.

49. anonymous

Ah, okay, you are using x^2 and applying a rule that the inverse would be negative in order to isolate the variables?

50. anonymous

oh

51. anonymous

Yeah so take like, (x^2)/(x^5). what do you get?

52. anonymous

But if you had x/x^2, you can't subtract the top x from the bottom, because you would have a 0 in the numerator, right?

53. anonymous

0/x^3?

54. anonymous

or x^-3/0? No, that can't be right

55. anonymous

Yes, which would make it one. Any number^0 is one. Like y^0 is one, z^0 is one... It's always one.

56. anonymous

You had it right the first time, but it cancels to one. So it would be 1/x^3.

57. anonymous

Okay, so if you take x^2/x^5 and turned the x^2 into x^-2 and subtract it by the denominator, you assume that there is a x^0 left over, and that equals 1?

58. anonymous

Yes, but that could get confusing quickly. Try this. Write out all of the x's on top and bottom. Then cross out as many as you can. Then explain to me what you did and what it means.

59. anonymous

But essentially yes, you are correct.

60. anonymous

x*x/x*x*x*x*x would be 0/x^3 or 1/x^3?

61. anonymous

It's one. You do NOT cancel to be zero. The only way to get zero would be to subtract something. Like, you have 4 quarters in a dollar. You cancel the 4 and you got 1 dollar right? You cant just have 4 quarters and then suddenly have nothing unless someone took the money from you, ie subtracted them. Does that make sense?

62. anonymous

Okay yes, that makes PERFECT sense :D I think I get it now :D

63. anonymous

64. anonymous

Alright, make sure you got this down. Make up a problem just like the one you originally asked and simplify it. I'll tell you if you did it right.

65. anonymous

And again, sorry for the trouble, but I think I completely get how to go about these problems now :D I can do alot of this math, but I missed a few basic concepts back in high school, and this helps me bridge those gaps :D

66. anonymous

Okay

67. anonymous

And no worries. Haha I'm Glad to help.

68. anonymous

Okay, making this up... x^-5y^9z^-2/x^3y^-5z^13, going to work it out now.

69. anonymous

so far, x^3y^14/x^5z^15

70. anonymous

y^14/x^2z^15 would be the final answer

71. anonymous

and if I multiplied everything by ^-2, that would make it x^4z^17/y^16?

72. anonymous

Almost perfect. But can you tell me what is wrong with the exponent on z and y?

73. anonymous

for which problem?

74. anonymous

the first one?

75. anonymous

do you multiply those exponents by 2 instead of adding 2 to them?

76. anonymous

oh yea, x^4z^34/y^32

77. anonymous

thats it, got it.

78. anonymous

Yes, you do multiply them, you only add when you are Multiplying. Ie,( x^2)^3 = x^6 (x^2)(x^2)= x^4

79. anonymous

$((3{}^{\wedge}-1)(x)(y{}^{\wedge}-4)(z{}^{\wedge}5)/(7x{}^{\wedge}-4)(y{}^{\wedge}2)(z)){}^{\wedge}-2$ $\left(\left(3^{-1}\right)(x)\left(y^{-4}\right)\left(z^5\right)/\left(7x^{-4}\right)\left(y^2\right)(z)\right)^{-2}$ $\left(\left(\frac{x}{3}\right)\frac{z^5}{y^4}/\frac{7}{x^4}\left(y^2 z\right)\right)^{-2}$ $\left(\frac{x z^5}{3 y^4}/\frac{7 y^2 z}{x^4}\right)^{-2}$ $\left(\frac{x^5 z^4}{21 y^6}\right)^{-2}$ $\frac{441 y^{12}}{x^{10} z^8}$

80. anonymous

Lol thanks, that's been established.

81. anonymous

This was done with Mathmatica 8. I over looked the reference to WolframAlpha.com in a prior post above. Sorry.