Question

If $\log_{e} (\frac{a + b}{2}) = \frac{1}{2} (\log_{e} a + \log_{e} b),$ then relation between a and b will be

Easy

Solution

$\log_{e} (\frac{a + b}{2}) = \frac{1}{2} (\log_{e} a + \log_{e} b)$
$= \frac{1}{2} \log_{e} (ab) = \log_{e} \sqrt{ab}$
$\Rightarrow \frac{a + b}{2} = \sqrt{ab} \Rightarrow a + b = 2 \sqrt{ab}$
$\Rightarrow {(\sqrt{a} - \sqrt{b})}^{2} = 0 \Rightarrow \sqrt{a} - \sqrt{b} = 0 \Rightarrow a = b$ .

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