Question

How do you solve this?

(x to the 6th power y to the 4th power) -1\2?

Answer 1

\[(x6 y ^4)-1\2\]

Answer 2

Are you solving for x or y or are you solving for when they = 0

Answer 3

I'm just trying to simply the equation and write the answer only using positive exponents

Answer 4

ok... use the powe of a power law

\[(x^a)^b = x^{a \times b}\]

a your problem can be split

\[(x^6)^{\frac{-1}{2}} \times (y^4)^{\frac{-1}{2}}\]

use the law above to simplify them

Answer 5

so multiply 6 by -1\2 and 4 by -1\2?

Answer 6

then you will be able to write them as fractions.

Answer 7

yep... thats the start

Answer 8

then negative powers indicate fractions

\[x^{-a} = \frac{1}{x^a}\]

Answer 9

so then would it be (x ^3) (y ^2)

Answer 10

well the powers after multiplying are

\[x^{-3}y^{-2}\]

Answer 11

now use the negative index law

Answer 12

ya I realized they should've been negative

Answer 13

so now do x ^-3 y ^-2

Answer 14

and then x to the -3 power = 1 over x cubed and the same thing for y, but with 2 for that one?

Answer 15

yes x^-3 you always want to flip to the bottom so 1/x^3 is the same thing but positive exponent.

Answer 16

so now that I've done that, what's next?

Answer 17

I guess according to him (1/x^3)*(1/y^2) you combine them so you should get something I think like 1/(X^3*y^2)

Answer 18

That actually makes sense a little bit. Thanks for trying to help :)

Answer 19

thats correct...

Answer 20

so I'm working another problem which is a ^3\4 and it needs to be turned into a radical. I have \[\sqrt a ^3 x \sqrt a ^4)\]

Answer 21

am i on the right track?