If the equation 2x+4y=2y+4x is solved for y in terms of x, is

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xlog2⁡1−2x

x+log2⁡1−2x

log2⁡1−2x

xlog2⁡2x+1

answer is B.

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Detailed Solution

22y−2y+2x1−2x=0 Putting 2y=t, we get t2−t+2x1−2x=0, where t1=2y1 and t2=2y2t1t2=2x1−2x2y1+y2=2x1−2xy1+y2=x+log2⁡1−2x

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