Question

The greatest integer less than $\frac{1}{{\log }_2\pi }+\frac{1}{{\log }_6\pi }$ is

Answer 1

Let $y=\frac{1}{{\log }_2\pi }+\frac{1}{{\log }_6\pi }$

Using ${\log }_{a}b=\frac{1}{{\log }_{b}a}$

$y={\log }_{\pi }2+{\log }_{\pi }6={\log }_{\pi }(2\times 6)\dots (\log a+\log b=\log (ab\left)\right)$

$y={\log }_{\pi }12$

${\pi }^y=12$

we know, ${\pi }^2<12<{\pi }^3\Rightarrow 2<y<3$

Thus, the greatest integer less than $y$ is $2.$