How do you expand #log_2(sqrt(32x^4)/root6(16y^3))#?

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Answer 1 · 2016-03-11T03:48:15

#log_2(sqrt(32x^4)/root(6)(16y^3)) = 0#

#(32x^4)^(1/2)/((16y^3)^(1/6)) = 1#

#((32x)^2)/(16y)^(1/2)= 1#

#(704x^2)/(sqrt(16y)) = 1#

#704x^2 = sqrt(16y)#

#(704x^2)^2 = (sqrt(16y))^2#

#495 616x^4= 16y#

#11264x^4 = y#

Hopefully this helps!