If log _k x cdot log _5 k = log _x 5,where k | vedclass

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(D) Given the equation: $\log _k x \cdot \log _5 k = \log _x 5$Using the change of base formula $\log _a b \cdot \log _b c = \log _a c$,we get:$\log _

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$5$ and $1/5$

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(D) Given the equation: $\log _k x \cdot \log _5 k = \log _x 5$Using the change of base formula $\log _a b \cdot \log _b c = \log _a c$,we get:$\log _5 x = \log _x 5$Let $y = \log _x 5$. Then $\log _5 x = 1/y$.So,$1/y = y$,which implies $y^2 = 1$.Therefore,$y = 1$ or $y = -1$.If $\log _x 5 = 1$,then $x^1 = 5$,so $x = 5$.If $\log _x 5 = -1$,then $x^{-1} = 5$,so $x = 1/5$.Thus,$x$ can be $5$ or $1/5$.

If $\log _k x \cdot \log _5 k = \log _x 5$,where $k \ne 1$ and $k > 0$,then $x$ is equal to:

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