Question

If $\frac{1}{lo{g}_3\pi }+\frac{1}{lo{g}_4\pi }>x$, then the biggest possible integer $x$ can take -

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is A

2

$\frac{1}{lo{g}_3\pi }+\frac{1}{lo{g}_4\pi }>x$
$\Rightarrow lo{g}_{\pi }3+lo{g}_{\pi }4>x\Rightarrow lo{g}_{\pi }12>x$
We know, ${\pi }^2<12<{\pi }^3$
$12>{\pi }^2\Rightarrow lo{g}_{\pi }12>lo{g}_{\pi }{\pi }^2\Rightarrow lo{g}_{\pi }12>2$
$\therefore xwillbe2$