Question

Rational Exponents
I need to write this expression in the simplest form:

(x^1/6 y^1/3)^-18

the -18 is loosing me please help

Answer 1

Is it minus 18 or times -18

Answer 2

its an exponent of -18

Answer 3

so what does a negative exponent mean do you know?

Answer 4

I know i divide but not sure how

Answer 5

Negative exponents mean it is in the denominator

Answer 6

So\[x^-2=1/x^2\]

Answer 7

the answer should be 1 / x3 y6 just not sure how to get it

Answer 8

What happens when you have a power raised to a power? You multiply the exponents right? So distribute the -18 and tell me what you get

Answer 9

I think I just got it, I need to take 1/6*18/1 =3 and 1/3*18/1 =6 so i would get the answer just not sure if i did it the right way.

Answer 10

Its not 18 its -18

Answer 11

So you should get -3 and -6

Answer 12

Now remember what i said earlier about negative exponents

Answer 13

right but my class was just told if we have a negative rational exponent you take the negative away by putting it all over 1 so it would be

1 / (x1/6 y1/3) exponent 18 would then = 1 / x3 y6

if I did this right the answer matches the book answer

Answer 14

yup that is it does it all make sense?

Answer 15

I think so now that - was throwing me for a while but I realized it just needed to cross multiply to get my answer. I think I was making it harder than it is and talking it out helped.

Answer 16

Im not quite understanding what you're writing. \[(x^{1/6}y^{1/3})^{-18}\]Is thatyour question?

Answer 17

yes that was the question

Answer 18

Oh, alright.

Answer 19

https://www.khanacademy.org/math/pre-algebra/exponents-radicals/exponent-properties/v/exponent-rules-part-2

Answer 20

In the future us (^) to denote raising something to a power

Answer 21

*use

Answer 22

how do i use that?

Answer 23

Shift and 6

Answer 24

ok thanks

Answer 25

So how I solved it: \[\large (x^{1/6}y^{1/3})^{-18} = \frac{1}{(x^{1/6}y^{1/3})^{18}} = \frac{1}{x^{(1/6) \cdot 18}y^{(1/3) \cdot 18}} =\frac{1}{x^3y^6} \]

Answer 26

yes, thank you

Answer 27

Sorry, I was also curious as to what your question meant because i was cnofused a bit by the notation.