જો ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$ તો $\sum {(1/x) = } $
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a
(a) \({a^{x - 1}} = bc \Rightarrow {a^x} = abc\)
(a) \({a^{x - 1}} = bc \Rightarrow {a^x} = abc\)
\(\therefore {a^x} = {b^y} = {c^z} = abc = k\,({\rm{say}})\)
\( \Rightarrow \)\(a = {k^{1/x}} \Rightarrow {1 \over x} = {\log _k}a\); \(\sum\limits_{}^{} {{1 \over x} = {{\log }_k}a + {{\log }_k}b + {{\log }_k}c} = {\log _k}abc = {\log _{abc}}abc = 1\).
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