Question

Without using tables, find that $\frac{1}{lo{g}_2\pi }+\frac{1}{lo{g}_5\pi }$ is always greater than

• A. $0$
• B. $1$
• C. $3/2$
• D. $2$

Answer 1

The correct option is D $2$

$\frac{1}{lo{g}_2\pi }+\frac{1}{lo{g}_5\pi }$

$=\frac{1}{\left(\frac{lo{g}_{\pi }\pi }{lo{g}_{\pi }2}\right)}+\frac{1}{\left(\frac{lo{g}_{\pi }\pi }{lo{g}_{\pi }5}\right)}$

$=\frac{lo{g}_{\pi }2}{lo{g}_{\pi }\pi }+\frac{lo{g}_{\pi }5}{lo{g}_{\pi }\pi }$

$=lo{g}_{\pi }2+lo{g}_{\pi }5$

$=lo{g}_{\pi }10$

$>lo{g}_{\pi }{\pi }^2$ sinc e$10>{\pi }^2$ and $lo{g}_{\pi }\left(x\right)$ is strictly increasing

$\therefore thegivenfunctionisalways>2$