Question

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

• A. $a=\frac{b}{2}$
• B. $a=b$
• C. $a=2b$
• D. none of these

Answer 1

The correct option is B

$a=b$

Explanation for the correct option

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

$\Rightarrow$ ${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}\left({\log }_{e}ab\right)$ $\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because {\log }_{a}m+{\log }_{a}n={\log }_a\mathrm{mn}\right]$

$\Rightarrow$ ${\log }_e\left(\frac{a+b}{2}\right)={\log }_e\sqrt{ab}$ $\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because {\mathrm{nlog}}_{a}m={\log }_a{m}^n\right]$

$\Rightarrow$ $\left(\frac{a+b}{2}\right)=\sqrt{ab}$ $\left[\because {\log }_{a}m={\log }_{a}n\Rightarrow m=n\right]$

$\Rightarrow$ $a+b=2\sqrt{ab}$

$\Rightarrow$${\left(\sqrt{a}\right)}^2-2\sqrt{a}\sqrt{b}+{\left(\sqrt{b}\right)}^2=0$

$\Rightarrow$ ${\left(\sqrt{a}-\sqrt{b}\right)}^2=0$ $\left[\because x^2-2\mathrm{xy}+y^2={\left(x-y\right)}^2\right]$

$\Rightarrow$ $\left(\sqrt{a}-\sqrt{b}\right)=0$

$\Rightarrow$ $\sqrt{a}=\sqrt{b}$

Squaring both sides we get

$\Rightarrow$ $a=b$

Hence, option (B) is the correct answer.