If x = {log _a}(bc),y = {log _b}(ca),z = {log | vedclass

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Question

(b) $x = {\log _a}bc$  $ \Rightarrow $ $1 + x = {\log _a}a + {\log _a}bc = {\log _a}abc$

$	herefore {(1 + x)^{ - 1}} = {\log _{abc}}a$

$

Vedclass · Accepted Answer

${(1 + x)^{ - 1}} + {(1 + y)^{ - 1}} + {(1 + z)^{ - 1}}$

Vedclass · Answer

(b) $x = {\log _a}bc$  $ \Rightarrow $ $1 + x = {\log _a}a + {\log _a}bc = {\log _a}abc$

$	herefore {(1 + x)^{ - 1}} = {\log _{abc}}a$

$	herefore {(1 + x)^{ - 1}} + {(1 + y)^{ - 1}} + {(1 + z)^{ - 1}} = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c$

$ = {\log _{abc}}abc = 1$.

If $x = {\log _a}(bc),y = {\log _b}(ca),z = {\log _c}(ab),$then which of the following is equal to $1$

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