જો ${a^x} = b,{b^y} = c,{c^z} = a,$ તો $xyz = . . . .$
Medium
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b
(b) \({a^x} = b \Rightarrow \) \(x\log a = \log b\)
(b) \({a^x} = b \Rightarrow \) \(x\log a = \log b\)
==> \(x = {{\log b} \over {\log a}} = {\log _a}b\)
Similarly \(y = {\log _b}c,\,z = {\log _c}a\)
\(\therefore xyz = {\log _a}b.{\log _b}c.{\log _c}a = 1\).
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