If log1/2xb−c=log1/2yc−a=log1/2za−b , then the value of xaybzc is equal to
A
12
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B
xyz
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C
(12)abc
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D
1
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Solution
The correct option is D1 Let log1/2xb−c=log1/2yc−a=log1/2za−b=k Now, alog1/2x+blog1/2y+clog1/2z=k[a(b−c)+b(c−a)+c(a−b)] ⇒alog1/2x+blog1/2y+clog1/2z=0 ⇒log1/2xaybzc=0[∵alogx=logxa&loga+logb+logc=log(abc)] ⇒xaybzc=1 Ans: D