Question

If $\log_{e}\left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_{e}a + \log_{e}b)$, 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

Answer: (A)

$a = b$

Answer 2

$\log_{e}\left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_{e}a + \log_{e}b) = \frac{1}{2}\log_{e}(ab) = \log_{e}\sqrt{ab}$

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