Question

$\text{If}{\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}({\log }_{e}a+{\log }_{e}b),$
$\text{then relation between}a\text{and}b\text{will be}$

Answer 1

Given:
${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}({\log }_{e}a+{\log }_{e}b)$

$\Rightarrow {\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}{\log }_e\left(ab\right)$
$\Rightarrow {\log }_e\left(\frac{a+b}{2}\right)={\log }_e\sqrt{ab}$
$\Rightarrow \frac{a+b}{2}=\sqrt{ab}$
$\Rightarrow a+b=2\sqrt{ab}$
$\Rightarrow a^{+}2ab+b^2=4ab$
$\Rightarrow a^{-}2ab+b^2=0$
$\Rightarrow (a-b{)}^2=0$
$\Rightarrow a=b$
Hence, The correct answer is option A