zer0 and negative integer exponents

[drenep212]

i need help with this problem. everyone seems to be getting the same answer as me and its wrong.
(2x^-2y)^-3 / (4x^2y^-1)^3

 

[tutor_joel]

Show us your work so we can see where you're having trouble. What answer did you get? you don't have to show every step. A couple steps will help and the answer as well.

It's a lot of work I know.

Is this the correct problem by the way?

$\displaystyle \frac{(2x^{-2y})^{-3}}{(4x^{2y^{-1}})^3}$

 

[drenep212]

Ive tried ever possible way including that..
the problem is.. the answer is 1/512. not possible to me but seems to be the answer all over the internet!
so (4 x^2 y^-1)^3 / (2 x^-2 y)^3
i solve integers first
4 x^6 y^-3 / 2 x^-6 y^3
everything pretty much cancels out. so how is the anser 1/ 512?

 

[drenep212]

x and y represent whole numbers

 

[tutor_joel]

1/512 is correct

So, you have

$\displaystyle \frac{(2x^{-2y})^{-3}}{(4x^{2y^{-1}})^3}$

Put the numerator in the denominator to change the sign to positive. You don't need to move the denominator as you did. The exponent is positive.

$\displaystyle \frac{1}{(2x^{-2y})^{3}(4x^{2y^{-1}})^3}$

Now, there are a couple ways to do the second step. I'll do the most obvious one. The cube distributes to every term. The 2 and the 4 as well. You missed this part.

$\displaystyle \frac{1}{(2^3x^{-2y*3})(4^3x^{3*2y^{-1}})} \;\;\; \;\;\; See\; how \;I \;got\; this?$

$\displaystyle \frac{1}{(8x^{-6y})(64x^{6y^{-1}})}$

you got it from here?

 

[drenep212]

no this is where your going wrong x and y represent whole numbers which need to be individually multiplied by the exponent. the y is not attached to the exponent of x they are whole numbers

 

[tutor_joel]

the y is not attached to the exponent of x they are whole numbers

I'm not sure what this means.

Believe me, this is how you solve the problem. I got an answer of 1/512. They need to state that x and y are whole numbers in order to do this kind of math.

You do not multiply the y^-1 exponent by 3.

$\displaystyle (4x^{2y^{-1}})^3 = (4^3)(x^{2y^{-1}})^3 =(4^3)x^{3(2y^{-1})}$

 

[tutor_joel]

Maybe a more obvious approach is

$\displaystyle \frac{1}{(2x^{-2y})^{3}(4x^{2y^{-1}})^3}$

$\displaystyle \frac{1}{(2^3)(4^3)(x^{-2y})^{3}(x^{2y^{-1}})^3}$

The two x terms cancel, yielding

$\displaystyle \frac{1}{(2^3)(4^3)}=\frac{1}{8^3}=\frac{1}{512}$

 

[drenep212]

then why isnt the exponent of x being multiplied by the exponent 3?
heres the problem.
(2 times the variable x to the power of -2 times the variable y) all to the power of -3
divided by
(4 times the variable x to the power of 2 times the variable y to the power of -1) all to the power of 3.
i dont understand why you keep putting y attached the the exponend of the variable x.

 

[tutor_joel]

The exponent of x is being multiplied by the exponent 3.

(2 times the variable x to the power of -2 times the variable y) all to the power of -3
divided by
(4 times the variable x to the power of 2 times the variable y to the power of -1) all to the power of 3.

So you have 4 terms here. Two in the numerator and two in the denominator. The terms are

2
x to the power of -2 times the variable y

4
x to the power of 2 times the variable y to the power of -1

now each term is cubed (raised to the power of 3), in the numerator it's negative, so we put it in the denominator to make it positive. I see you have that part. so you have:

2 cubed
(x to the power of -2 times the variable y) cubed

4 cubed
(x to the power of 2 times the variable y to the power of -1) cubed

now, the rules of math state that when raising a term to some power (3 for this problem) - for a term that has an exponent already, you multiply the exponents together. The exponent of the two x terms are

(the power of 2 times the variable y to the power of -1) times 3, yielding 6y^(-1) for the exponent of x

and

(the power of -2 times the variable y) times 3, yielding -6y for the exponent of the other x term

 

[mmm4444bot]

?

Joel, I'm thinking that the OP is trying to say that y is not an exponent.

The end result is the same, however.

There are at least four different ways to simplify the given expression (which the OP apparently mistyped as two different expressions).

Here's another way, that uses properties of exponents, as well as the fact that 4 is 2^2.

$\displaystyle \frac{(2 \ x^{-2} \ y)^{-3}}{(2^2 \ x^2 \ y^{-1})^3}$

?

$\displaystyle \frac{2^{-3} \ x^6 \ y^{-3}}{2^6 \ x^6 \ y^{-3}}$

?
$\displaystyle \frac{2^{-3}}{2^6} \cdot \frac{x^6}{x^6} \cdot \frac{y^{-3}}{y^{-3}}$

?

$\displaystyle 2^{-3 - 6} \cdot 1 \cdot 1$

?

$\displaystyle 2^{-9}$

?

$\displaystyle \frac{1}{512}$

?

 

[tutor_joel]

Re:

?

Joel, I'm thinking that the OP is trying to say that y is not an exponent.

The end result is the same, however.

There are at least four different ways to simplify the given expression (which the OP apparently mistyped as two different expressions).

Here's another way, that uses the fact that 4 is 2^2.

$\displaystyle \frac{(2 \ x^{-2} \ y)^{-3}}{(2^2 \ x^2 \ y^{-1})^3}$

$\displaystyle \frac{2^{-3} \ x^6 \ y^{-3}}{2^6 \ x^6 \ y^{-3}}$

$\displaystyle \frac{2^{-3}}{2^6} \cdot \frac{x^6}{x^6} \cdot \frac{y^{-3}}{y^{-3}}$

$\displaystyle 2^{-3 - 6} \cdot 1 \cdot 1$

$\displaystyle 2^{-9}$

$\displaystyle \frac{1}{512}$

oic, Duh.

 

[tutor_joel]

i need help with this problem. everyone seems to be getting the same answer as me and its wrong.
(2x^-2y)^-3 / (4x^2y^-1)^3

Should be

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

for your future reference. I got confused by your notation and when I got the right answer, I thought I had originally wrote it correctly. Sorry for the confusion.

Hey, on the bright side, you're still learning stuff.

technically

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

but who's gonna be that nit picky - besides a machine that is

 

[drenep212]

this is how the book is giving it to me.. exactly as ive put it.. but where i was going wrong was that i was assuming the denom problem would be switched with the numerator (recipricol) and then trying to get 512.. problem was in deviding it gets reduced to 1/8 instead of 1/512 because you have to mulitply 8 and 64. trying to see it step by step wasnt helping me cuz you were making Y an exponent... and confusing the living daylights out of me! but i see the picture now..

 

[mmm4444bot]

?

who's gonna be that nit picky

Me! :twisted:

Actually, any way that is not ambiguous works well.

(Personally, I have a dislike for adjacent operators, too, so I would put grouping symbols around negative exponents.)

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

 

[mmm4444bot]

this is how the book is giving it to me.. exactly as ive put it

The publisher used a typewriter, instead of professional typography? :shock:

That's not good at all, for a math text.

 

[drenep212]

well its their 5th edition. and all the other problems look like that too.. heres another one i dont understand the wording to.. u up for it?!

 

[tutor_joel]

Let's see it.

What's the ISBN number for that book, I've got to see this. Publishers make bank on textbooks and they have a tendency to write really bad books. It's all about pretty pictures nowadays.

 

[soroban]

Hello, drenep212!

$\displaystyle \text{Simplify: }\;\frac{\left(2x^{-2}y\right)^{-3}} {\left(4x^2y^{-1}\right)^3}$

$\displaystyle \frac{\left(2x^{-2}y\right)^{-3}}{\left(4x^2y^{-1}\right)^3} \;=\; \frac{2^{-3} \left(x^{-2}\right)^{-3}\left(y^{-3}\right)} {4^3\left(x^2\right)^3\left(y^{-1}\right)^3}$

. . . . . . . . \(\displaystyle = \;\frac{2^{-3}\left(x^6\right)\left(y^{-3}\right)} {4^3\left(x^6)\left(y^{-3}\right)}\)

. . . . . . . . \(\displaystyle = \;\frac{2^{-3}\rlap{////}\left(x^6\right)\, \rlap{/////} \left(y^{-3}\right)} {4^3\rlap{////}\left(x^6)\,\rlap{/////}\left(y^{-3}\right)}\)

. . . . . . . . $\displaystyle =\;\frac{2^{-3}}{4^3} \;\;=\;\;\frac{1}{2^3}\cdot\frac{1}{4^3} \;\;=\;\;\frac{1}{8}\cdot\frac{1}{64} \;\;=\;\;\frac{1}{512}$