Integer Exponents

Question

Can I have assistance with three problems?

1. (3x^2y-4) (4x^-3y)

2. (15x^-1y^2/5xy^-4)-1

3. 24x^-1y^-2/6x^-4y^3

Answer 1

Remember these rules of exponents.

a^maⁿ = a^m+n, a^m/aⁿ = a^m-n, and a^-n = 1/aⁿ.  Also (a/b)^-1 = b/a.

Regroup the terms before applying these rules, and you will make everything a little easier.

1. (3x^2y-4) (4x^-3y) = 3*4 x²x^-3 y^-4y = 12*x^-1*y^-3 = 12/(xy³).

Try the others using the rules this way.  If you're still having trouble, I or another will be happy to offer further assistance.

Answer 2

Hello Taylor,

These all are exponent problems.

1. (3x²y^-4)(4x^-3y)

In this problem we are multiplying. First we'll multiply the coefficients and then we'll multiply the variables with bases x and y.

     3.4.x².x^-3.y^-4.y

   =12x^2-3.y^-4+1       (formula x^m.xⁿ = x^m+n)

   = 12x^-1y^-3

   = 12/xy3 (answer)  (x^-1 = 1/x and y^-3 = 1/y³)

 

2.   (15x^-1y²/5xy^-4)^-1

First we'll take care inside the bracket and solve. Take care of coefficient and then variables.

   = [(15/5).(x^-1/x).(y²/y^-4)]^-1

   = [3(1/x¹⁺¹)(y²⁺⁴)]^-1        (formula used x^-1 = 1/x, x^m/xⁿ = x^m-n) 

    = [3(1/x²)(y⁶)]^-1

    = (3^-1.y^6.-1)/x^2.-1                (power to power formula (x^m)ⁿ = x^mn)

   = 3^-1.y^-6/x^-2

   = x²/3y⁶ (answer)

 

3.  24x^-1y^-2/6x^-4y³

   = (24/6)(x^-1/x^-4)(y^-2/y³)

   = 4(x^-1+4)(1/y³⁺²)    (formula used 1/x^-1 = x, x^-1 = 1/x, x^m/xⁿ = x^m-n)

   = 4x³/y⁵ (answer)

Hope this helps.

Answer 3

Please clarify where the exponents begin and end for each x term.

If problem #1 looks like this: (3x²y-4)(4x^-3y) then here's the explanation.

Because we have two terms in the first parenthese, we need to distribute the second parenthese to each of those terms. Thus we have:

3x²y*4x^-3y-4*4x^-3y

Now to simpligy we multiply the constants with constants, x's with x's and y's with y's.

12x²*x^-3*y*y-16x^-3y. The exponents for the x's add together as per exponential laws (x^a*x^b=x^a+b). Same for the y's:

12x^-1y²-16x^-3y

The x's will need to be brough down because they have a negative exponent (x^-a=1/x^a).

12y²/x-16y/x³ And lastly we can make a common denominator by multiplying the left term by x² on both top and bottom and combine the two terms:

(12y²x²-16y)/x³ And that's our simplest answer.

Hope this helped!

Answer 4

1. (3x²y^-4)(4x^-3y) = (3x4)(x^(2-3))(y^(-4+1)) = 12x^-1y^-3

2. (15x^-1y² / 5x y^-4)^-1 = ( (15/5)(x^(-1-1))(y⁽²⁺⁴⁾) )^-1 = ( 3 x^-2y⁶ )^-1 = (1/3) x^(-2x-1)y^(6x-1) = (1/3) x²y^-6

3. (24x^-1y-2)/(6x^-4y³) = (24/6)(x^(-1+4))(y^(-2-3)) = 4x³y^-5

you can use general formula as below

1/x^a = x^-a , x^a time x^b = x^(a+b) , (x^a)^b = x^ab