If a, b, c are distinct positive numbers, eac | vedclass

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Question

(a) $[{\log _b}a.{\log _c}a - {\log _a}a] + [{\log _a}b.{\log _c}b - {\log _b}b]$
$ + [{\log _a}c{\log _b}c - {\log _c}c] = 0$

==> ${1 \over {\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a) $[{\log _b}a.{\log _c}a - {\log _a}a] + [{\log _a}b.{\log _c}b - {\log _b}b]$
$ + [{\log _a}c{\log _b}c - {\log _c}c] = 0$

==> ${1 \over {\ln a.\ln b.\ln c}}[{(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c] = 0$

==> ${(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c = 0$

==> $\ln a + \ln b + \ln c = 0$

==> $\ln (abc) = ln 1$, $[{a^3} + {b^3} + {c^3} - 3abc = 0$

==> $a + b + c = 0]$,

$	herefore abc = 1$.

If $a, b, c$ are distinct positive numbers, each different from $1$, such that $[{\log _b}a{\log _c}a - {\log _a}a] + [{\log _a}b{\log _c}b - {\log _b}b]$ $ + [{\log _a}c{\log _b}c - {\log _c}c] = 0,$ then $abc =$

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