Write without negative exponents: (2x^-3y^4)^-2?

(2x⁻³y⁴)⁻² = (2)⁻²(x⁻³)⁻²(y⁴)⁻²
= (2⁻²)(x⁺⁶)(y⁻⁸)
= (1/2²)(x⁶)(1/y⁸)
= x⁶/(4y⁸)

That is not a question.

Also - do you know who is going to get the good jobs when you graduate? The kids who know how to do these things. Better study harder if you want a good life.

4x^9y^8

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

(1/4) x^6 y^(-8)

x^6
----------
4 y^8

Okay i will try to explain everything as much as i can, so in exchange just take your time and read it.If you can't right now then leave and come back later, just know that i will try to explain everything so that you won't need any help again for this kind of problem. So here we go.
To do this kinds of problem you will need to know this three little formulas:

first:
3 = 3^1, or 4 = 4^1, or 100 = 100^1 or 2505250 = 2505250^1 etc... ---hope you got it, natural numbers are always to the power of " 1 "

second:
(x^1 )^ 2 = [ x^ (1 * 2 )] ----multiplying the power give the answer. example: ( 2^1)^ 2 = 2^(1*2) = 2^2 or ( 2^ 1)^ -2 = 2^(1*-2) = 2^ -2

and, third:
x^ -1 = 1 / x^1 or x^ -2 = 1/ x^ 2 or x^ -100 = 1 / x^ 100 etc....
--- hope you got it, a numerator with a " negative power " can be switch to a denominator and be of " positive power "
this only work on multiplication and division, so be careful.( don't use it on addition or subtraction )

So now let's go back to your question.

(2x^ -3 y^4)^ -2 ---- this is the normal problem

[ (2^ 1)(x^ -3)(y^ 4) ] ^ -2 ---- this is how their powers are distributed don't forget " 2 " is a natural number so 2 = 2^ 1.

[ (2^(1*-2)) (x^ (-3 *-2)) (y^(4 * -2) ] ---- multiply the power " -2 " by the power of each one of them.

[ (2^ -2)(x^ 6)(y^ -8) ] ----- simple multiplication

1/4 *(x^ 6)(y^ -8) ---- " 2^ -2 = 1 / 2^ 2 = 1 /4 ", look back at the "second and last formula "that i wrote in the beginning.

[ 1* (x^ 6)(y^ -8) ] / 4 ---- multiply "4" by the denominator of the others (witch are "1 ")

[ 1* x^6 ( 1/ y^ 8) ] / 4 ---- " y ^ -8 = 1 / y^8 ", look back at " the last formula "that i wrote in the beginning."

(1* x^6 * 1) / 4y^8 ---- just rewriting

( x^ 6 ) / ( 4y^ 8 ) ---- here is your answer

As you know, a work that can't be prove is just a guess and because this answer is not a guess, then we have to prove it so here we go:
To prove this, just choose two numbers that you want and switch them to the "x " and " y " in the question that you asked in the beginning witch was: " (2x^-3y^4)^-2 " and in also our final answer which was: " ( x^ 6 ) / ( 4y^ 8 ) ".
If both Answers are the same then the answer is right however if they don't, then the answer is wrong.

For this i will choose " 2 for x " and " 1 for y " ( don't forget you can choose any number but they have to be different)
so:
x = 2
y = 1
put into the first equation we will have:

(2(2)^-3 *(1)^4)^-2 = ( 2* 1/8 * 1 )^ -2 = 1/4 * 64 *1 = 64 / 4 = 16

now put into the second equation we will have:

( 2^ 6 ) / ( 4(1)^ 8 ) = 64 / ( 4 * 1 ) = 64 / 4 = 16 ---- did you saw this it's "16 " yeah "16 " again!!!! do you know what that mean, yep!! the answer is right.!!!!!!!

In conclusion it's 100% sure that your answer is:
" ( x^ 6 ) / ( 4y^ 8 ) "

Hope that i helped bye.

(2x^-3y^4)^-2
= (2^-2)[x^(-3 * -2)][y^(4 * -2)]
= [1/(2^2)][x^6][y^-8]
= [1/4][x^6][1/(y^8)]
= x^6/(4y^8)

Distribute ^-2 to the terms in the parenthesis.
= (2x^-3y^4)^-2
= 2^-2x^6y^-8

We can never have a negative exponent in our final answer. To convert this into positive, get its reciprocal. In this case, we'll bring down 2^-2, and y^-8 to the denominator.

= x^6 / (2)^2 y^8
= x^6 / 4 * y^8
= x^6 / 4y^8

Hope this helps.

Hope this helps.

yeah...i dunno what the heck you were talking about...

(2x^-3y^4)^-2=
(3 (2 x^3+x+1))/x^2