What is #(2x^0 * 2x^-2)/((2y^2)^4)#?

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Answer 1 · 2016-03-24T07:10:02

#1/(4x^2y^8)#

Before we simplify this remember the identities

#a^0=1#, #a^(-m)=1/a^m#, #(a^m)^n=a^(mn)# and #(ab)^m=a^m*b^m#

Using these, #(2x^0*2x^(-2))/((2y^2)^4)# can be expressed as

#(2*1*2*(1/(x^2)))/((2^4*(y^2)^4)#

(as #x^0=1#, #x^(-2)=1/x^2#, #(2y^2)^4=2^4*y^8#)

Now this can be further simplified as

#(4)/(x^2*16*y^8)# or #1/(4x^2y^8)#