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 $\log _b a = \frac{\ln a}{\ln b}$,we can rewrite the

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(D) Given the equation: $\log _k x \cdot \log _5 k = \log _x 5$.Using the change of base formula $\log _b a = \frac{\ln a}{\ln b}$,we can rewrite the expression as:$\frac{\ln x}{\ln k} \cdot \frac{\ln k}{\ln 5} = \frac{\ln 5}{\ln x}$.Simplifying the left side by canceling $\ln k$:$\frac{\ln x}{\ln 5} = \frac{\ln 5}{\ln x}$.Cross-multiplying gives:$(\ln x)^2 = (\ln 5)^2$.Taking the square root on both sides:$\ln x = \ln 5$ or $\ln x = -\ln 5$.This implies $x = 5$ or $x = 5^{-1} = \frac{1}{5}$.

If $\log _k x \cdot \log _5 k = \log _x 5$,where $k \neq 1$ and $k > 0$,then the value of $x$ is:

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