Question

The value of $\frac{1}{{\log }_3\pi }+\frac{1}{{\log }_4\pi }$ is

• A. $2$
• B. Less than $2$
• C. More than $2$
• D. None of these

Answer 1

Answer: (C)

More than $2$

We have,

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

We know that

${\log }_{a}x=\frac{{\log }_{e}x}{{\log }_{e}a}$

Therefore,

$=\frac{1}{\frac{{\log }_e\pi }{{\log }_{e}3}}+\frac{1}{\frac{{\log }_e\pi }{{\log }_{e}4}}$

$=\frac{{\log }_{e}3}{{\log }_e\pi }+\frac{{\log }_{e}4}{{\log }_e\pi }$

$=\frac{{\log }_{e}3+{\log }_{e}4}{{\log }_e\pi }$

We know that

${\log }_{e}m+{\log }_{e}n={\log }_e\left(mn\right)$

Therefore,

$=\frac{{\log }_{e}12}{{\log }_e\pi }$

$=\frac{2.484}{1.144}$

$=2.17$

Hence, the value is more than $2$.