Question

simplify using only positive exponents: (4x^3 y^-2)^0/(2x^-4 y^3)^-3 (3x^5 y)^-2

Answer 1

can you draw it?

Answer 2

\[\huge \frac{(4x^3y^{-2})^0}{(2x^-4y^3)^{-3}(3x^5y)^{-2}}\]
It's a little hard to read, is this what the problem looks like?

Answer 3

yes except its 2x^-4

Answer 4

the numerator is just 1 :P

Answer 5

because its raised to the 0 power right?

Answer 6

\[\huge \frac{(4x^3y^{-2})^0}{(2x^{-4}y^3)^{-3}(3x^5y)^{-2}}\]Ah sorry bout that :) Formatting error.

Answer 7

no worries! :]

Answer 8

thats awesome how you did that, and clear text!

Answer 9

yup laylam

Answer 10

Yes because of the 0 power.

When you square something, you multiply it by itself, so it becomes squared.
When you raise something to the 1st power, you do nothing to it.

When you raise something to the 0th power, you divide it by itself, giving you 1.

Answer 11

\[\huge \frac{1}{(2x^{-4}y^3)^{-3}(3x^5y)^{-2}}\]Just need to take care of the mess on the bottom now :)

Answer 12

The negative power tells us that we can FLIP the term if we change it to positive.

Example:\[\huge \frac{1}{x^{-2}} \qquad = \qquad x^2\]See how I changed it to a positive and brought it up top?

Answer 13

okay great! So since (2x^-4 y^3)^-3 is raised to the -3, do we cube that?

Answer 14

yup!

Answer 15

We want to first bring it up top, since it's a negative in the denominator :)
But then, yes, we'll be able to cube it after that.

Answer 16

okay so it would be (2x^-4 y^3)^3 (3x^5y)^2 at the top?

Answer 17

Yes good good c:

Answer 18

yay!

Answer 19

Remember how to apply the exponents to terms in a bracket set?
You will apply it to EVERYTHING in the brackets!

Answer 20

so it would be (2x^-12 y^9)^3 (3x^10y)^2

Answer 21

Well, a couple things to be careful about.

When you apply the exponent, you shouldn't have it floating outside anymore.
Because the way you have it written now, it looks like you still need to apply the exponent to it.

Also, don't forget to apply the exponent to CONSTANT terms, not just variables.
(Hint hint: the 2 and 3)

Answer 22

oh I see, so like this?

(6x^-12 3y^9) (6x^10 2y)

Answer 23

From this,\[\huge (2x^{-4}y^3)^{3}(3x^5y)^{2}\]
You got,
\[\huge (\color{orangered}{6}x^{-12} \color{orangered}{3}y^9)(\color{orangered}{6}x^{10}\color{orangered}{y})\]
The red are the parts we need to fix.

Answer 24

The constant in the first bracket should be,\[\huge 2^3\]Which is not 6 :O

Answer 25

is the x still there?

Answer 26

Anything I left in black was correct, so don't worry about the X's :)

Answer 27

are you just attaching the exponent to each variable and not multiplying it?

Answer 28

\[\huge (2x^{-4}y^3)^3 \qquad = \qquad 2^3(x^{-4})^3(y^3)^3\]We cube eeeeeverything. See how that works? :D

Answer 29

I'm confused :/ then for (y^3)^3 why isn't it going to be (3y^9)?

Answer 30

If you wanted to, you could think of the y term like this,\[\huge (1\cdot y^3)^3\]When we cube this, we'll get,\[\huge 1^3\cdot y^9 \quad =\quad y^9\]
I'm not really sure where your 3 is coming from.

Answer 31

ohh I see! I just multiplied the exponent to the y

Answer 32

so would it be (2x^-12 y^9) (3x^10 y^2)

Answer 33

You have your x's and y's correct.

The constants we still need to fix though.

Answer 34

You didn't apply the exponent to your constants in front.

Answer 35

(2^3 x^-12 y^9) (3^2 x^10 y^2)

Answer 36

Yes good :) now simplify the constants down silly! heh

Answer 37

ah see thats where I was confused when you said "apply the exponent to the constant" haha thats where I multiplied -_-

Answer 38

Hah XD my bad, maybe i could have explained that better.

Answer 39

no worries! Okay so it's (8 x^-12 y^9)(9 x^10 y^2)

Answer 40

Yes looks good. and from here we can simplify further I'm afraid!! :(

Answer 41

:o

Answer 42

We can multiply out the brackets :OOO

Answer 43

multiplying the x's and the y's together?

Answer 44

Luckily, each bracket has only one TERM, so no FOIL nonsense.

Answer 45

Yes :) Remember what to do with the exponents when you multiply bases?

Answer 46

phew!

Answer 47

they get added!

Answer 48

Yesss c:

Answer 49

72x^-3 2y^11 !!!!

Answer 50

Uh ohhhhhhhhhhhhhhhhhhhhhhh,\[\huge x^{-12}\cdot x^{10}=x^{\color{orangered}{-3}}\]

Answer 51

Red means bad! XD lol

Answer 52

but what about 8*9?!

Answer 53

\[\huge 8 \cdot 9=72\]Ok that part looks good. Why is there a 2 in front of the y? :\

Answer 54

so thats right 72x^-3? oh I thought because there are two y's :/ so then its just

Answer 55

72x^-3 y^11 ?

Answer 56

Why are you getting -3 on the X?

Answer 57

-2!! my bad lol

Answer 58

Yay team \:D/

Answer 59

omg haha silly mistakes!

Answer 60

but thank you so much!!!

Answer 61

:D

Answer 62

Yah easy to make silly mistakes :D long problem. heh