Question

Let ${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}({\log }_{e}a+{\log }_{e}b)$. If $a=4$, then the value of $b$ would be

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is A $4$

Given, ${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}({\log }_{e}a+{\log }_{e}b)=\frac{1}{2}{\log }_e\left(ab\right),[\because \log x+\log y=\log (xy\left)\right]$
$\Rightarrow {\log }_e\left(\frac{a+b}{2}\right)=\left({\log }_e\sqrt{ab}\right)[\because x\log a=\log a^x]$
$\Rightarrow \frac{a+b}{2}=\sqrt{ab}$
$\Rightarrow a+b-2\sqrt{ab}=0$
$\Rightarrow {(\sqrt{a}-\sqrt{b})}^2=0$
$\Rightarrow a=b$
Given $a=4\Rightarrow b=4$