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(x^4y^3)^(-2)/(x^3y^2)^(-4) =
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Bunuel
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(x^4y^3)^(-2)/(x^3y^2)^(-4) = [#permalink] New post  04 Apr 2018, 22:40
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\(\frac{(x^4y^3)^{(-2)}}{(x^3y^2)^{(-4)}}=\)


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

B. \(xy\)

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

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

E. \(x^4y^2\)
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KSBGC
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Re: (x^4y^3)^(-2)/(x^3y^2)^(-4) = [#permalink] New post  04 Apr 2018, 22:54
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. \(x^4y^2\)


we know that if we have negative inverse sign we can easily avoid it changing their numerator and denominator position.

(x^3. y^2)^4/(x^4. y^3)^2

=x^12. y^8 / x^8. y^6
=x^4.y^2

which is option E.
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generis
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Re: (x^4y^3)^(-2)/(x^3y^2)^(-4) = [#permalink] New post  05 Apr 2018, 11:17
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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 E

Simpler 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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Re: (x^4y^3)^(-2)/(x^3y^2)^(-4) = [#permalink] New post  23 Dec 2018, 06:39
We can easily solve this with: (x^a)^b = x^(a*b) and (x^a)/(x^b) = x^(a-b)
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Abhishek009
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Re: (x^4y^3)^(-2)/(x^3y^2)^(-4) = [#permalink] New post  23 Dec 2018, 07:19
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. \(x^4y^2\)


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

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

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

= \(x^4y^2\), Answer must be (E)
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Re: (x^4y^3)^(-2)/(x^3y^2)^(-4) = [#permalink]
23 Dec 2018, 07:19
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