If log_e left( frac{a + b}{2} right) = frac{1 | vedclass

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(A) Given $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$.Using the property $\log_e x + \log_e y = \log_e(xy)$,we get $\fr

Vedclass · Accepted Answer

$a = b$

Vedclass · Answer

(A) Given $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$.Using the property $\log_e x + \log_e y = \log_e(xy)$,we get $\frac{1}{2}(\log_e a + \log_e b) = \frac{1}{2} \log_e(ab) = \log_e(ab)^{1/2} = \log_e \sqrt{ab}$.So,$\log_e \left( \frac{a + b}{2} \right) = \log_e \sqrt{ab}$.Taking the exponential on both sides,we get $\frac{a + b}{2} = \sqrt{ab}$.$a + b = 2\sqrt{ab} \implies a + b - 2\sqrt{ab} = 0$.$(\sqrt{a} - \sqrt{b})^2 = 0$.Therefore,$\sqrt{a} = \sqrt{b}$,which implies $a = b$.

If $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$,then the relationship between $a$ and $b$ is:

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