Question

Simplify the given expression using only positive exponents. Then complete the statements that follow. [(X^2y^3)^-1/(x^-2y^2z)^2]^2

The exponent on x is___
The exponent on y is___
The exponent on a is____

Answer 1

As

`\:\left((\left(x^2y^3\right)^(-1))/(\left(x^(-2)y^2z\right)^2)\right)^2=x^4y^(-14)z^(-4)`

• The exponent on x is 4

• The exponent on y is -14

• The exponent on z is -4

Explanation:

Given the expression

`\left[(\left(x^2y^3\right)^(-1))/(\left(x^(-2)y^2z\right)^2)\right]^2`

`\mathrm{Apply\:exponent\:rule}:\quad \left((a)/(b)\right)^c=(a^c)/(b^c)`

`\left((\left(x^2y^3\right)^(-1))/(\left(x^(-2)y^2z\right)^2)\right)^2=(\left(\left(x^2y^3\right)^(-1)\right)^2)/(\left(\left(x^(-2)y^2z\right)^2\right)^2)`

`=(\left(\left(x^2y^3\right)^(-1)\right)^2)/(\left(\left(x^(-2)y^2z\right)^2\right)^2)`

as

`\mathrm{Apply\:exponent\:rule}:\quad \:a^(-1)=(1)/(a)`

so the expression becomes

`=((1)/(x^4y^6))/(\left(\left(x^(-2)y^2z\right)^2\right)^2)` ∵
`\left(x^2y^3\right)^(-1)=(1)/(x^2y^3)`

as

`\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n`

so the expression becomes

`=((1)/(x^4y^6))/((y^8z^4)/(x^8))` ∵
`\left(x^(-2)y^2z\right)^2=(y^4z^2)/(x^4)`

as

`\mathrm{Divide\:fractions}:\quad ((a)/(b))/((c)/(d))=(a\cdot \:d)/(b\cdot \:c)`

so the expression becomes

`=(1\cdot \:x^8)/(x^4y^6y^8z^4)`

`=(x^8)/(x^4y^6y^8z^4)`

as

`\mathrm{Apply\:exponent\:rule}:\quad (x^a)/(x^b)=x^(a-b)`

so the expression becomes

`=(x^(8-4))/(y^8y^6z^4)`

`=(x^4)/(y^8y^6z^4)`

`=(x^4)/(y^(14)z^4)`

as

`\mathrm{Apply\:exponent\:rule}:\quad \:a^(-1)=(1)/(a)`

so the expression becomes

`=x^4y^(-14)z^(-4)`

Therefore,

• The exponent on x is 4

• The exponent on y is -14

• The exponent on z is -4