Bunuel wrote:

\(\frac{(x^4y^3)^{(-2)}}{(x^3y^2)^{(-4)}}=\)

A. \(x^{-4}y^{-2}\)

B. \(xy\)

C. \(x^3y^3\)

D. \(x^2y^2\)

E. \(x^4y^2\)

Three steps:

\(\frac{(x^4y^3)^{(-2)}}{(x^3y^2)^{(-4)}} = \frac{(x^{-8}y^{-6})}{(x^{-12}y^{-8})}=\)

\(x^{-8-(-12)}y^{-6-(-8)} = x^4y^2\)

Answer ESimpler steps.

Invert the fraction, change exponents' signs*

\(\frac{(x^4y^3)^{(-2)}}{(x^3y^2)^{(-4)}}=\frac{(x^3y^2)^{(4)}}{(x^4y^3)^{(2)}}\)

Distribute the exponent:

\((a^{m})^{n}=a^{m*n}=a^{mn}\)\(\frac{(x^3y^2)^{(4)}}{(x^4y^3)^{(2)}}=\frac{x^{12}y^8}{x^8y^6}\)

Divide: keep the base, subtract the exponent

\(\frac{x^{12}y^8}{x^8y^6}= x^{(12-8)}y^{(8-6)}= x^4y^2\)

Answer E*A term with a negative exponent in the numerator or denominator can be moved to the opposite side of the fraction, i.e., to denominator or numerator, respectively. Change the sign of the exponent.

\(\frac{4^{-2}}{2^{-2}}=\frac{(\frac{1}{4^2})}{(\frac{1}{2^2})}=(\frac{1}{4^2}*\frac{2^2}{1})=(\frac{1}{16}*4)=\frac{4}{16}\)

Easier:\(\frac{4^{-2}}{2^{-2}}=\frac{2^2}{4^2}=\frac{4}{16}\)
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