Respuesta :
Answer:
As
[tex]\:\left(\frac{\left(x^2y^3\right)^{-1}}{\left(x^{-2}y^2z\right)^2}\right)^2=x^4y^{-14}z^{-4}[/tex]
- The exponent on x is 4
- The exponent on y is -14
- The exponent on z is -4
Step-by-step explanation:
Given the expression
[tex]\left[\frac{\left(x^2y^3\right)^{-1}}{\left(x^{-2}y^2z\right)^2}\right]^2[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}[/tex]
[tex]\left(\frac{\left(x^2y^3\right)^{-1}}{\left(x^{-2}y^2z\right)^2}\right)^2=\frac{\left(\left(x^2y^3\right)^{-1}\right)^2}{\left(\left(x^{-2}y^2z\right)^2\right)^2}[/tex]
[tex]=\frac{\left(\left(x^2y^3\right)^{-1}\right)^2}{\left(\left(x^{-2}y^2z\right)^2\right)^2}[/tex]
as
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{-1}=\frac{1}{a}[/tex]
so the expression becomes
[tex]=\frac{\frac{1}{x^4y^6}}{\left(\left(x^{-2}y^2z\right)^2\right)^2}[/tex] ∵ [tex]\left(x^2y^3\right)^{-1}=\frac{1}{x^2y^3}[/tex]
as
[tex]\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n[/tex]
so the expression becomes
[tex]=\frac{\frac{1}{x^4y^6}}{\frac{y^8z^4}{x^8}}[/tex] ∵ [tex]\left(x^{-2}y^2z\right)^2=\frac{y^4z^2}{x^4}[/tex]
as
[tex]\mathrm{Divide\:fractions}:\quad \frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot \:d}{b\cdot \:c}[/tex]
so the expression becomes
[tex]=\frac{1\cdot \:x^8}{x^4y^6y^8z^4}[/tex]
[tex]=\frac{x^8}{x^4y^6y^8z^4}[/tex]
as
[tex]\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}[/tex]
so the expression becomes
[tex]=\frac{x^{8-4}}{y^8y^6z^4}[/tex]
[tex]=\frac{x^4}{y^8y^6z^4}[/tex]
[tex]=\frac{x^4}{y^{14}z^4}[/tex]
as
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{-1}=\frac{1}{a}[/tex]
so the expression becomes
[tex]=x^4y^{-14}z^{-4}[/tex]
Therefore,
- The exponent on x is 4
- The exponent on y is -14
- The exponent on z is -4