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If a, b, c are positive real numbers then `(1)/("log"_(a)bc_(+1)) + (1)/("log"_(b)ca_(+1)) +(1)/("log"_(c)ab_(+1))=` |
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Answer» Correct Answer - B We have, `(1)/("log"_(a)bc_(+1)) + (1)/("log"_(b)ca_(+1)) +(1)/("log"_(c)ab_(+1))` `=(1)/("log"_(a)bc + "log"_(a)a) + (1)/("log"_(b)ca + "log"_(b)b) + (1)/("log"_(c)ab + "log"_(c)c)` `=(1)/("log"_(a)abc) +(1)/("log"_(b)abc) + (1)/("log"_(c)abc)` ` = "log"_(abc) a + "log"_(abc) b + "log"_(abc) c = "log"_(abc)abc = 1` |
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