Question

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

• A. $a=b$
• B. $a=\frac{b}{2}$
• C. $2a=b$
• D. $a=\frac{b}{3}$

Answer 1

The correct option is A $a=b$

$lo{g}_e\left(\frac{a+b}{2}\right)=\frac{1}{2}(lo{g}_{e}a+lo{g}_{e}b)$
R.H.S $=\frac{1}{2}lo{g}_e\left(ab\right)=lo{g}_e\sqrt{ab}$
$\Rightarrow \frac{a+b}{2}=\sqrt{ab}\Rightarrow a+b=2\sqrt{ab}$
$\Rightarrow (\sqrt{a}-\sqrt{b}{)}^2=0\Rightarrow \sqrt{a}-\sqrt{b}=0\Rightarrow a=b$