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Basic of Logarithms
medium
If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be
A
$2$
B
$3$
C
$3.5$
D
$\pi $
Solution
(a) ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x$
$ \Rightarrow $ ${\log _\pi }3 + {\log _\pi }4 > x$$ \Rightarrow $ ${\log _\pi }12 > x$
${\pi ^2} < 12 < {\pi ^3}$
$\therefore 12 > {\pi ^2}$;
$\therefore {\log _\pi }12 > {\log _\pi }{\pi ^2}$
i.e., ${\log _\pi }12 > 2$;
$\therefore x$ will be $2$.
Standard 11
Mathematics
