જો ${\log _k}x.\,{\log _5}k = {\log _x}5,k \ne 1,k > 0$ તો $x = . . . .$
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d
(d) \({\log _k}x.{\log _5}k = {\log _x}5\) \( \Rightarrow \) \({\log _5}x = {\log _x}5\)
(d) \({\log _k}x.{\log _5}k = {\log _x}5\) \( \Rightarrow \) \({\log _5}x = {\log _x}5\)
\( \Rightarrow \) \({\log _x}5 = {1 \over {{{\log }_x}5}}\) \(\Rightarrow \) \(1 = - 2B = 0\)
\( \Rightarrow \) \({\log _x}5 = \pm 1\)
\( \Rightarrow \) \({x^{ \pm \,1}} = 5\) \( \Rightarrow \) \(x = 5,\,{1 \over 5}\).
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