Question

Okay, lets try this again...

[(3^-1)(x)(y^-4)(z^5)/(7x^-4)(y^2)(z)]^-2

Answer 1

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

Answer 2

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

Answer 3

Yes I did, but now I got stuck

Answer 4

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

Answer 5

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.

Answer 6

oh yea, thats right

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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?

Answer 12

*I did fix the z

Answer 13

sorry it's messy

Answer 14

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

Answer 15

eh, 49 rather

Answer 16

Yes. Everything gets flipped because they all become negative. Then you multiply all exponents by 2 and square the 21.

You should get:
\[x ^{10} z ^{8} \div 441 y ^{12}\]

Answer 17

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

Answer 18

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

Answer 19

Yeah sorry I typed it backwards. Yours is right.

Answer 20

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?

Answer 21

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?

Answer 22

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.

Answer 23

Rain, I don't think that is right.

Answer 24

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

Answer 25

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.

Answer 26

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.

Answer 27

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?

Answer 28

Here, let me explain another way.

How would you simplify \[x ^{2} \div x\] ?

Answer 29

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

Answer 30

That would be just x

Answer 31

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.

Answer 32

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?

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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.

Answer 37

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

Answer 38

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

Answer 39

I think you got it right this time.

Answer 40

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.

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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?

Answer 45

x/x=x?

Answer 46

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.

Answer 47

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

Answer 48

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

Answer 49

oh

Answer 50

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

Answer 51

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?

Answer 52

0/x^3?

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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?

Answer 57

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.

Answer 58

But essentially yes, you are correct.

Answer 59

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

Answer 60

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?

Answer 61

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

Answer 62

I just wanted to be sure I completely understood how you were going about this.

Answer 63

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.

Answer 64

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

Answer 65

Okay

Answer 66

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

Answer 67

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

Answer 68

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

Answer 69

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

Answer 70

and if I multiplied everything by ^-2, that would make it

x^4z^17/y^16?

Answer 71

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

Answer 72

for which problem?

Answer 73

the first one?

Answer 74

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

Answer 75

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

Answer 76

thats it, got it.

Answer 77

Yes, you do multiply them, you only add when you are Multiplying.

Ie,( x^2)^3 = x^6

(x^2)(x^2)= x^4

Answer 78

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

Answer 79

Lol thanks, that's been established.

Answer 80

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