How do you simplify ${(\frac{2 m}{n^{2}})}^{4}$ $((2m)/ n^2)^4$ ?

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Answer 1 · 2015-07-13T00:51:03

${(\frac{2 m}{n^{2}})}^{4} = \frac{16 m^{4}}{n^{8}}$ $((2m)/n^2)^4 = (16m^4)/n^8$

You must apply the outside exponent to everything inside the parentheses.

${(\frac{2 m}{n^{2}})}^{4} = \frac{2^{4} m^{4}}{{(n^{2})}^{4}}$ $((2m)/n^2)^4 = (2^4m^4)/(n^2)^4$

We have to repeat the procedure with the denominator.

$\frac{2^{4} m^{4}}{{(n^{2})}^{4}} = \frac{2^{4} m^{4}}{n^{8}}$ $(2^4m^4)/(n^2)^4 = (2^4m^4)/n^8$

So

${(\frac{2 m}{n^{2}})}^{4} = \frac{2^{4} m^{4}}{n^{8}} = \frac{16 m^{4}}{n^{8}}$ $((2m)/n^2)^4 = (2^4m^4)/n^8=(16m^4)/n^8$

Answer 2 · 2015-07-13T01:05:51

The answer is $\frac{16 m^{4}}{n^{8}}$ $(16m^4)/n^8$ .

${(\frac{2 m}{n^{2}})}^{4}$ $((2m)/(n^2))^4$

${(\frac{a}{b})}^{x} = \frac{a^{x}}{b^{x}}$ $((a)/(b))^x=(a^x)/(b^x)$

$\frac{{(2 m)}^{4}}{{(n^{2})}^{4}}$ $(2m)^4/(n^2)^4$

${(a^{x})}^{y} = a^{x \cdot y}$ $(a^x)^y=a^(x*y)$

$\frac{{(2 m)}^{4}}{n^{2 \cdot 4}}$ $(2m)^4/(n^(2*4)$ =

$\frac{{(2 m)}^{4}}{n^{8}}$ $(2m)^4/n^8$

${(a b)}^{x} = a^{x} b^{y}$ $(ab)^x=a^xb^y$

$\frac{2^{4} m^{4}}{n^{8}}$ $(2^4m^4)/n^8$ =

$\frac{16 m^{4}}{n^{8}}$ $(16m^4)/n^8$

How do you simplify ((2m)/ n^2)^4(2mn2)4((2m)/ n^2)^4?