Question

(x^-2y^-4x^3)^-2 First remove the parenthesis now we use the ab = 1/ab property x^-2 = 1/x^2 now we have a positive exponent x^2. Next y^-4 = 1/y^4 now we have three positive exponents now we need to make the negative 2 positive. -2= 1/-2
now we have x^2y^4x^3 add the like terms x^2+ x^3 = x^5
(x^5y^4)^2

Answer 1

Can you help me with this @misssunshinexxoxo

Answer 2

@SolomonZelman?

Answer 3

\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~y^{-4}~x^3\right)^{-2} }\) like this?

Answer 4

Yeah. Brb five minutes or less sorry making lunch.

Answer 5

:)

Answer 6

Ok, I will tell you another property.....

\(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\)

Answer 7

this property, can be applied to the terms with an x.
lets do a little replacement, just to visualize it better:

\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~y^{-4}~\cdot~x^3\right)^{-2} }\)

\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\)

Answer 8

so, you can apply the property I told you just now to the first 2 terms in this expression.

Answer 9

Back Okay.

Answer 10

ok, take your time to read, and apply the property i have provided...

Answer 11

So we don't change the terms into positive exponents?

Answer 12

we will do that later, but for now we just want to simplify inside the parenthesis (using this new property i told you).....

Answer 13

\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\)

(this is your expression - the order of how you multiply doesn't matter)

----------------------------------------------------
Knowing that:
\(\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}\)

then,
\(\displaystyle \Large {\rm x}^\color{red}{\rm \LARGE -2}\times{\rm w}^\color{blue}{\rm \LARGE 3}={\rm x}^{?}\)

Answer 14

By the bases being different i can't add them together right. so i guess x would be still x^-2

Answer 15

you can apply it to the " x\(^{-2}\) • x\(^3\) " part (where the base is same, the base is x)

Answer 16

we would just have x^1 which we don't count so we just have x.

Answer 17

Sorry thought i already enter it forgot i had to click on post. >~< @SolomonZelman

Answer 18

yes, x^1 is correct.

Answer 19

And x^1 is the same exact thing as x

(besides the fact that you waste extra space and time when writing ^1 there)

Answer 20

we had:
\(\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }\)

our new expression (that is equivalent to the previous one, of course) is:
\(\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }\)

Answer 21

now, apply the property of

\(a^{-b}\)=\(\dfrac{1}{a^b}\)

to the " y\(^{-4}\) " part

Answer 22

Okay y^-4 =1/y^4 @SolomonZelman

Answer 23

yes

Answer 24

So how do i solve for -2 then?

Answer 25

our expression:
\( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }\)

is going to become:
\( \LARGE\color{black}{ \displaystyle \large \left(x~\cdot~\dfrac{1}{y^{4}}\right)^{-2} }\)

\( \LARGE\color{black}{ \displaystyle \large \left(\dfrac{x}{y^{4}}\right)^{-2} }\)

Answer 26

Okay, so we don't change two?

Answer 27

I will give you another propery (which can is derived from \(a^{-b}\)=1 /\(a^b\))

the property is:

\(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\)

Answer 28

Text - correct:
`which can *be* derived...`

Answer 29

Knowing that:

\(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }\)

then, therefore:

\(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2} }\)

Answer 30

so far so good?

Answer 31

Yeah just make the reciprocal and it turn the negative into a positive.

Answer 32

Thanks for your help again. I appreciate it. @SolomonZelman

Answer 33

this is not it

Answer 34

There's more DX

Answer 35

\(\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)

Answer 36

yes, this one more step...

Answer 37

Alright.

Answer 38

\((\)y\(^4\)\()\)\(\large^{^2}\) = y\(^{4~\times~ 4}\) = y\(^?\)

Answer 39

y^16

Answer 40

yes, y\(^{16}\)/x\(^2\)

Answer 41

everything makes sense?

Answer 42

Yeah so far do i divide by x or i'm just writing it as y^16 over x^2

Answer 43

you had

\(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }\)

Answer 44

so we multiply the exponent y^4 with 2 and we get 16/x^2 Yeah it makes sense :)

Answer 45

applying the rule:

\(\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{\rm A}}{\color{red}{\rm B}}\right)^{\Large w}=\dfrac{\left(\color{blue}{\rm A}\right)^w}{\left(\color{red}{\rm B}\right)^w} }\)

Answer 46

and then after we applied this rule above, we simplify on the top.

Answer 47

Which gives the result of: y\(^{16}\)/x\(^2\)

Answer 48

and that is it. can't do anything with this....

Answer 49

Alright thanks @SolomonZelman

Answer 50

yw, gtg.