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# (x^4y^3)^(-2)/(x^3y^2)^(-4) =

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Joined: 02 Sep 2009
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04 Apr 2018, 23:40
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86% (01:02) correct 14% (01:40) wrong based on 102 sessions

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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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04 Apr 2018, 23:54
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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05 Apr 2018, 12:17
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:

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

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

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

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

*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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Joined: 03 Nov 2018
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23 Dec 2018, 07:39
We can easily solve this with: (x^a)^b = x^(a*b) and (x^a)/(x^b) = x^(a-b)
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23 Dec 2018, 08:19
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, 08:19
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