Sequence And Series Question 599
Question: If $ {\log_{x}}a,\ {a^{x/2}} $ and $ {\log_{b}}x $ are in G.P., then $ x= $
Options:
A) $ -\log ({\log_{b}}a) $
B) $ -{\log_{a}}({\log_{a}}b) $
C) $ {\log_{a}}({\log_{e}}a)-{\log_{a}}({\log_{e}}b) $
D) $ {\log_{a}}({\log_{e}}b)-{\log_{a}}({\log_{e}}a) $
Show Answer
Answer:
Correct Answer: C
Solution:
Obviously $ {{({a^{x/2}})}^{2}}={\log_{x}}a\ .\ {\log_{b}}x={\log_{b}}a $
$ \Rightarrow $ $ a^{x}={\log_{b}}a $
$ \Rightarrow $ $ x={\log_{a}}({\log_{b}}a) $
$ \Rightarrow $ $ x={\log_{a}}( \frac{{\log_{e}}a}{{\log_{e}}b} )={\log_{a}}({\log_{e}}a)-{\log_{a}}({\log_{e}}b) $ .
BETA
