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If a, b, c are positive real numbers, then `(1)/("log"_(ab)abc) + (1)/("log"_(bc)abc) + (1)/("log"_(ca)abc) =`

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If a, b, c are positive real numbers, then
`(1)/("log"_(ab)abc) + (1)/("log"_(bc)abc) + (1)/("log"_(ca)abc) =`
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Correct Answer - C
We have,
`(1)/("log"_(ab)abc) + (1)/("log"_(bc)abc) + (1)/("log"_(ca)abc) =`
` = "log"_(abc)ab + "log"_(abc)bc + "log"_(abc)ca`
` = "log"_(abc) (ab xx bc xx ca) = "log"_(abc) (abc)^(2) = 2"log"_(abc) abc = 2`
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