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8. Sequences and Series
easy
यदि ${\log _x}a,\;{a^{x/2}}$ व ${\log _b}x$ गुणोत्तर श्रेणी में हों, तब $x =$
A
$ - \log ({\log _b}a)$
B
$ - {\log _a}({\log _a}b)$
C
${\log _a}({\log _e}a) - {\log _a}({\log _e}b)$
D
${\log _a}({\log _e}b) - {\log _a}({\log _e}a)$
Solution
(c) स्पष्टत:, ${({a^{x/2}})^2} = {\log _x}a\;.\;{\log _b}x = {\log _b}a$
$ \Rightarrow $ ${a^x} = {\log _b}a$
$ \Rightarrow $ $x = {\log _a}({\log _b}a)$
$ \Rightarrow $ $x = {\log _a}\left( {\frac{{{{\log }_e}a}}{{{{\log }_e}b}}} \right) = {\log _a}({\log _e}a) – {\log _a}({\log _e}b)$.
Standard 11
Mathematics
