If log _x a, a^{x/2} and log _b x are in G.P. | vedclass

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(C) Given that $\log _x a, a^{x/2}, \log _b x$ are in $G.P.$By the property of $G.P.$,the square of the middle term is equal to the product of the ext

Vedclass · Accepted Answer

$\log _a(\log _e a) - \log _a(\log _e b)$

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(C) Given that $\log _x a, a^{x/2}, \log _b x$ are in $G.P.$By the property of $G.P.$,the square of the middle term is equal to the product of the extremes:$(a^{x/2})^2 = (\log _x a) \cdot (\log _b x)$$a^x = \frac{\log a}{\log x} \cdot \frac{\log x}{\log b}$$a^x = \frac{\log a}{\log b} = \log _b a$Taking $\log _a$ on both sides:$x = \log _a(\log _b a)$Using the change of base formula $\log _b a = \frac{\log _e a}{\log _e b}$:$x = \log _a\left( \frac{\log _e a}{\log _e b} \right)$Using the logarithmic property $\log (m/n) = \log m - \log n$:$x = \log _a(\log _e a) - \log _a(\log _e b)$.

If $\log _x a, a^{x/2}$ and $\log _b x$ are in $G.P.$,then $x = $

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