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Question

If $$\cfrac { \log _{ 1/2 }{ x }  }{ b-c } =\cfrac { \log _{ 1/2 }{ y }  }{ c-a } =\cfrac { \log _{ 1/2 }{ z }  }{ a-b } $$ , then the value of $${ x }^{ a }{ y }^{ b }{ z }^{ c }$$ 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 D $$1$$
Let  $$\cfrac { \log _{ 1/2 }{ x }  }{ b-c } =\cfrac { \log _{ 1/2 }{ y }  }{ c-a } =\cfrac { \log _{ 1/2 }{ z }  }{ a-b } =k$$
Now, $$a\log _{ 1/2 }{ x } +b\log _{ 1/2 }{ y } +c\log _{ 1/2 }{ z } =k\left[ a\left( b-c \right) +b\left( c-a \right) +c\left( a-b \right)  \right] $$
$$\Rightarrow a\log _{ 1/2 }{ x } +b\log _{ 1/2 }{ y } +c\log _{ 1/2 }{ z } =0$$
$$\Rightarrow \log _{ 1/2 }{ { x }^{ a }{ y }^{ b }{ z }^{ c } } =0 \quad [\because a\log x=\log x^a   \quad \text{&} \quad   \log a+\log b+\log c=\log(abc)]$$
$$\Rightarrow { x }^{ a }{ y }^{ b }{ z }^{ c }=1$$
Ans: D

Mathematics

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