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((m^-3)(n^2)^-4)/((m^2)(n^-3))^2

 

A quick question, how do you get (m^8/n^2) out of this? I have no idea what to approach where. It's a little infuriating.

 May 7, 2017
 #1
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+1

Simplify the following:
(1)/(((n^2)/(m^3))^4 ((m^2)/(n^3))^2)

Multiply each exponent in m^2/n^3 by 2:
(1)/((n^2/m^3)^4 (m^(2×2))/((n^3)^2))

2×2 = 4:
(1)/((n^2/m^3)^4 m^4/(n^3)^2)

Multiply exponents. (n^3)^2 = n^(3×2):
(1)/((n^2/m^3)^4 m^4/n^(3×2))

3×2 = 6:
(1)/((n^2/m^3)^4 m^4/n^6)

(n^2/m^3)^(-4) = (m^3/n^2)^4:
((m^3/n^2)^4)/(m^4/n^6)

Multiply each exponent in m^3/n^2 by 4:
((m^(4×3))/((n^2)^4))/(m^4/n^6)

4×3 = 12:
(m^12/(n^2)^4)/(m^4/n^6)

Multiply exponents. (n^2)^4 = n^(2×4):
(m^12/n^(2×4))/(m^4/n^6)

2×4 = 8:
(m^12/n^8)/(m^4/n^6)

Multiply the numerator by the reciprocal of the denominator, (m^12/n^8)/(m^4/n^6) = m^12/n^8×n^6/m^4:
(m^12 n^6)/(n^8 m^4)

Combine powers. (m^12 n^6)/(n^8 m^4) = m^(12 - 4) n^(6 - 8):
m^12 - 4 n^6 - 8

12 - 4 = 8:
m^8 n^(6 - 8)

6 - 8 = -2:
Answer: | m^8 n^-2

 May 7, 2017
 #2
avatar+7347 
+4

This might be a little bit easier to see :)

 

\(\large \frac{(m^{-3}n^2)^{-4}}{(m^2n^{-3})^2} \\~\\ =\frac{m^{12}n^{-8}}{m^4n^{-6}} \\~\\ =m^{12-4}n^{-8--6} \\~\\ =m^{8}n^{-2} \\~\\ =\frac{m^{8}}{n^{2}}\)

.
 May 7, 2017

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