How do you simplify #(\frac { 2m ^ { - 1} n ^ { 4} \cdot 3m ^ { 6} n ^ { 2} } { 4m ^ { 7} n ^ { 6} } ) ^ { 4}#?

1 Answer
EZ as pi
Nov 6, 2016

#81/(16m^6)#

Explanation:

#((2m^-1 n^4*3m^6n^2)/(4m^7n^6))^4#

It does not matter which of the laws of indices are applied first, but as there is quite a lot happening inside the bracket it might be a good idea to simplify the fraction first before removing the bracket.

#((6m^5 n^6)/(4m^7n^6))^4" "larr# simplify the numerator - add indices

#(3/(2m^2))^4" "larr# simplify the fraction - subtract indices

#3^4/(2^4m^6)" "larr# raise to a power, multiply indices

#81/(16m^6)#

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