Question

If $p=\frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }+1$, then

• A. $2.5<p<3$
• B. $p>3$
• C. $1.5<p<2$
• D. $2<p<2.5$

Answer 1

Answer: (B)

$p>3$

Given, $p=\frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }+1$

Now, $p={\log }_{\pi }3+{\log }_{\pi }4+1$

$=\frac{\log 3}{\log \pi }+\frac{\log 4}{\log \pi }+1$
$=\frac{\log 3+\log 4+\log \pi }{\log \pi }=\frac{\log (3\times 4\times \pi )}{\log \pi }$

$=\frac{\log 12\pi }{\log \pi }={\log }_{\pi }12\pi =3.1707$

$\therefore p>3$.