Question

Prove that: $\frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }>2$

Answer 1

Consider,

$\frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }$

$={\log }_{\pi }3+{\log }_{\pi }4$ ($\because {\log }_{a}b=\frac{1}{{\log }_{b}a}$)

$={\log }_{\pi }(3\times 4)$ ($\because \log mn=\log m+\log n$)

$\Rightarrow \frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }={\log }_{\pi }12$ .....(1)

As $12>{\pi }^2$

$\Rightarrow {\log }_{\pi }12>{\log }_{\pi }{\pi }^2$

$\Rightarrow {\log }_{\pi }12>2$ ($\because {\log }_{a}a=1$)

So, using this in (1), we get

$\frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }>2$