Question

If ${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}\left[{\log }_e\left(a\right)+{\log }_e\left(b\right)\right]$then

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

Answer 1

The correct option is A

$a=b$

Explanation for the correct option:

${\log }_e\left(\frac{a+b}{2}\right)=\frac{1}{2}\left[{\log }_e\left(a\right)+{\log }_e\left(b\right)\right]$

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

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

$\Rightarrow$ ${\log }_e\left(\frac{a+b}{2}\right)={\log }_e\left(\sqrt{ab}\right)$

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

Squaring on both sides

$\Rightarrow$ $\frac{{\left(a+b\right)}^2}{4}=ab$

$\Rightarrow$ $a^2+2ab+b^2=4ab$

$\Rightarrow$ $a^2-2ab+b^2=0$

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

$\Rightarrow$ $a-b=0$

$\Rightarrow$ $a=b$

Hence, option(A) i.e. $a=b$ is the correct answer.