Question #d5bad

1 Answer
Oct 15, 2017

x/(2y^4)x2y4

Explanation:

Given, (4x^-2. y^8)^(-1/2)(4x2.y8)12

rArr 4^(-1/2). (x^-2)^(-1/2). (y^8)^(-1/2)412.(x2)12.(y8)12

rArr (2^2)^(-1/2). x^[(-2)(-1/2)]. y^[8(-1/2)](22)12.x(2)(12).y8(12) [As (a^m)^n = a^(mn)](am)n=amn]

rArr 2^-1. x^1. y^(-4)21.x1.y4 [ As a^-1 = 1/a]a1=1a]

rArr x/(2y^4)x2y4