If {1 over {{{log }_3}pi }} + {1 over {{{log | vedclass

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Question

(a) ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x$

$ \Rightarrow $ ${\log _\pi }3 + {\log _\pi }4 > x$$ \Rightarrow $ ${\log

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(a) ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x$ $ \Rightarrow $ ${\log _\pi }3 + {\log _\pi }4 > x$$ \Rightarrow $ ${\log _\pi }12 > x$ ${\pi ^2} < 12 < {\pi ^3}$ $ herefore 12 > {\pi ^2}$; $ herefore {\log _\pi }12 > {\log _\pi }{\pi ^2}$ i.e., ${\log _\pi }12 > 2$; $ herefore x$ will be $2$.

If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be

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