If x = {log _b}a,,,y = {log _c}b,,,,z = {log | vedclass

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(b) We have $xyz = {\log _b}a 	imes {\log _c}b 	imes {\log _a}c$

$ = {{{{\log }_e}a} \over {{{\log }_e}b}} 	imes {{{{\log }_e}b} \over {{{\log }

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$1$

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(b) We have $xyz = {\log _b}a 	imes {\log _c}b 	imes {\log _a}c$

$ = {{{{\log }_e}a} \over {{{\log }_e}b}} 	imes {{{{\log }_e}b} \over {{{\log }_e}c}} 	imes {{{{\log }_e}c} \over {{{\log }_e}a}} = 1$.

If $x = {\log _b}a,\,\,y = {\log _c}b,\,\,\,z = {\log _a}c$, then $xyz$ is

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