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:\(\frac{4^{-2}}{2^{-2}}=\frac{2^2}{4^2}=\frac{4}{16}\)
_________________
SC Butler has resumed! Get
two SC questions to practice, whose links you can find by date,
here.Choose life.