If {a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - | vedclass

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(a) ${a^{x - 1}} = bc \Rightarrow {a^x} = abc$

$	herefore {a^x} = {b^y} = {c^z} = abc = k\,({m{say}})$

$ \Rightarrow $$a = {k^{1/x}} \Rightar

Vedclass · Accepted Answer

$1$

Vedclass · Answer

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

If ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$then $\sum {(1/x) = } $

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