If log_x a, a^{x/2} and log_b x are in G.P.,t | vedclass

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(C) Given that $\log_x a, a^{x/2}, \log_b x$ are in $G.P.$Therefore,$(a^{x/2})^2 = (\log_x a) \cdot (\log_b x)$.Using the change of base formula $\log

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$\log_a(\log_e a) - \log_a(\log_e b)$

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(C) Given that $\log_x a, a^{x/2}, \log_b x$ are in $G.P.$Therefore,$(a^{x/2})^2 = (\log_x a) \cdot (\log_b x)$.Using the change of base formula $\log_x a \cdot \log_b x = \frac{\log a}{\log x} \cdot \frac{\log x}{\log b} = \frac{\log a}{\log b} = \log_b a$.So,$a^x = \log_b a$.Taking $\log_a$ on both sides,we get $x = \log_a(\log_b a)$.Using the change of base formula for $\log_b a = \frac{\log_e a}{\log_e b}$,we have $x = \log_a\left(\frac{\log_e a}{\log_e b}\right)$.By the property of logarithms,$\log_a\left(\frac{M}{N}\right) = \log_a M - \log_a N$.Thus,$x = \log_a(\log_e a) - \log_a(\log_e b)$.

If $\log_x a, a^{x/2}$ and $\log_b x$ are in $G.P.$,then $x = $

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