x(y+z−x)=logyy(z+x−y)=logzz(x+y−z)=k1
logx=k(xy+xz−x2)⋯(1)
logy=k(yz+xy−y2)⋯(2)
logz=k(zx+yz−z2)⋯(3)
(2)×x⟹xlogy=k(xyz+x2y−xy2)⋯(4)
(3)×y⟹ylogx=k(xy2+xyz−x2y)⋯(5)
(4)+(5)⟹ylogx+xlogy=2k(xyz)⟹logxyyx=2k(xyz)
Similarly logzyyz=2k(xyz),logxzzx=2
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Answers & Comments
x(y+z−x)=logyy(z+x−y)=logzz(x+y−z)=k1
logx=k(xy+xz−x2)⋯(1)
logy=k(yz+xy−y2)⋯(2)
logz=k(zx+yz−z2)⋯(3)
(2)×x⟹xlogy=k(xyz+x2y−xy2)⋯(4)
(3)×y⟹ylogx=k(xy2+xyz−x2y)⋯(5)
(4)+(5)⟹ylogx+xlogy=2k(xyz)⟹logxyyx=2k(xyz)
Similarly logzyyz=2k(xyz),logxzzx=2