How do you simplify #(2a^-1b^-2*-ba^-3)/(2a^0b^4)^-1#?

1 Answer
EZ as pi
Sep 13, 2016

#(4b^3)/(a^4)#

Explanation:

Recall: #x^-m = 1/x^m#

#(2a^-1b^-2xxba^-3)/color(red)((2a^0b^4)^-1)" " larr# get rid of negative indices

=#(2b xxcolor(red)(2cancela^0b^4))/(ab^2a^3)" "larr# simplify top and bottom

=#(4b^5)/(a^4b^2)" "larr# subtract indices of like bases

=#(4b^3)/(a^4)#

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