Answer:
m^2/n
Step-by-step explanation:
The following laws of exponent are useful to this problem:
[tex] \displaystyle \large{ {(mn)}^{b} = {m}^{b} {n}^{b} } \\ \displaystyle \large{ {n}^{ - b} = \frac{1}{ {n}^{b} } }[/tex]
We are given the expression:
[tex] \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } }[/tex]
Use the first law of exponent above.
[tex] \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{(8)( \frac{1}{4}) } {n}^{ ( - 4)( \frac{1}{4}) } } \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } {n}^{ - 1} }[/tex]
Make sure to recall the important necesscary fundamental math such as operation with negative numbers/integers, basic division, fraction, etc.
From the expression, apply the second law of exponent to n^-1.
[tex] \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n}^{1} } )} \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n} } )} \\ [/tex]
Multiply m in.
[tex] \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = \frac{ {m}^{2} }{ {n} } }[/tex]
Thus the answer is m^2/n.
