Q. If $\log_{e}\left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_{e}{a + \log_{e}b})$, then
A) $a = b$
B) $a = \frac{b}{2}$
C) $2a = b$
D) $a = \frac{b}{3}$
Explanation
(a)
Given, $\log_{e}{\left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_{e}{a + \log_{e}b})}$
$\Rightarrow \therefore\ \ \ \frac{a + b}{2} = \sqrt{ab}$
$\Rightarrow a + b - 2\sqrt{ab} = 0$
$\Rightarrow \ \ \ \ \sqrt{a} = \sqrt{b}$
$\Rightarrow a = b$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.