If _e ( a + b 2 ) = 1 2 ( _e a + _e b), then relation between a a… - Vidyadip

MCQ

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

✓
$a = b$
B
$a = {b \over 2}$
C
$2a = b$
D
$a = {b \over 3}$

✓

Answer

Correct option: A.

$a = b$

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

$ = {1 \over 2}{\log _e}(ab) = {\log _e}\sqrt {ab} $

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

==> $\,a = b$.

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