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Question: If \(\log_{x}a,a^{x/2}\)and \(\log_{b}x\) are in G.P. then \(x\) is equal to...

If log⁡xa,ax/2\log_{x}a,a^{x/2}and log⁡bx\log_{b}x are in G.P. then xx is equal to

A

log⁡a(log⁡ba)\log_{a}\left( \log_{b}a \right)

B

log⁡a(log⁡ea)−log⁡a(log⁡eb)\log_{a}\left( \log_{e}a \right) - \log_{a}\left( \log_{e}b \right)

C

−log⁡a(log⁡ab)- \log_{a}\left( \log_{a}b \right)

D

log⁡a(log⁡eb)−log⁡a(log⁡ea)\log_{a}\left( \log_{e}b \right) - \log_{a}\left( \log_{e}a \right)

Answer

log⁡a(log⁡ea)−log⁡a(log⁡eb)\log_{a}\left( \log_{e}a \right) - \log_{a}\left( \log_{e}b \right)

Explanation

Solution

As log⁡xa,ax/2,log⁡bx\log_{x}a,a^{x/2},\log_{b}xare in G.P.

(ax/2)2=log⁡xa.log⁡bx\left( a^{x/2} \right)^{2} = \log_{x}a.\log_{b}x

⇒ax=log⁡alog⁡x.log⁡xlog⁡b=log⁡alog⁡b=log⁡baa^{x} = \frac{\log a}{\log x}.\frac{\log x}{\log b} = \frac{\log a}{\log b} = \log_{b}a

⇒ x=log⁡a(log⁡ba)=log⁡a(log⁡ea)−log⁡a(log⁡eb)x = \log_{a}\left( \log_{b}a \right) = \log_{a}\left( \log_{e}a \right) - \log_{a}\left( \log_{e}b \right)