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3. The ordered pair (x, y) satisfying the equation x2=1+6log4 y and y2=2xy+22x+1 are (x1, y1) and (x2, y2), then find the value of  log 2 x 1   x 2   y 1   y 2 = ?

x2=1+6 log4yNote:y>0y2=2xy+22x+1y2=2xy+2.22xy2=2xy+2.2x2Let t=2xy2=ty+2t22t2+yt-y2=02t2+2yt-yt-y2=02tt+y-yt+y=0t+y2t-y=0t+y=0 or 2t-y=0y=-t or y=2tBut y>0 and also t>0 so y-tHence, y=2ty=2.2xy=2x+1x2=1+6 log4 2x+1x2=1+6 log4 22x+12x2=1+6 log4 4x+12Use logabm=m logabx2=1+6x+12 log4 4x2=1+6x+12x2=1+3x+3x2-3x-4=0x2-4x+x-4xx-4+x-4=0x-4x+1=0x=4 or x=-1at x=4, y=24+1=25=32at x=-1, y=2-1+1=20=1Hence two ordered pair are 4,32 and-1,1log2x1 x2 y1 y2=log2 4.32.1.-1=log2 4.32.1.-1=log2 22.25=log2 27Use logabm=m logab=7log2 2=7

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