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a
If a2+4b2=12ab, then log(a+2b)=12(loga+logb)
b
If logxb-c=logyc-a=logza-b then xa·yb·zc=abc
c
1logxyxyz+1logyzxyz+1logzxxyz=2
d
All are correct
answer is C.
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Detailed Solution
(1) log(a+2b)=12loga+2b2 =12loga2+4b2+4ab =12log(12ab+4ab)=12log24⋅ab =12(4log2+loga+logb)(2) Let logxb−c=logyc−a=logza−b=k⇒logx=k(b−c),logy=k(c−a)logz=k(a−b) xa⋅yb⋅zc=pk[a(b−c))+b(c−a)+c(a−b)]=pk(0)=1(3) Given expression is =logxyzxy+logxyzyz+logxyzzx =logxyz(xy⋅yz⋅zx)=logxyzx2⋅y2⋅z2 =2logxyz(xyz)=2×1=2