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Question

If log(x+y6)=12(logx+logy), show that xy+yx=34

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Solution

log(x+y6)=12(logx+logy)
2log(x+y6)=(logx+logy)
log(x+y6)2=log(xy)
logx+logy=log(xy)
(x+y)236=xy
x2+y2+2xy=36xy
x2+y2+2xy36xy=0
x2+y234xy=0
x2+y2=34xy
x2xy+y2xy=34
xy+yx=34

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