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

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(A) Given the equation: $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$ Using the property $\log_e x + \log_e y = \log_e (x

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$a = b$

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(A) Given the equation: $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2}(\log_e a + \log_e b)$ Using the property $\log_e x + \log_e y = \log_e (xy)$,we get: $\log_e \left( \frac{a + b}{2} \right) = \frac{1}{2} \log_e (ab)$ Using the property $n \log_e x = \log_e (x^n)$,we get: $\log_e \left( \frac{a + b}{2} \right) = \log_e (ab)^{1/2} = \log_e \sqrt{ab}$ Taking the exponential of both sides: $\frac{a + b}{2} = \sqrt{ab}$ $a + b = 2\sqrt{ab}$ $a + b - 2\sqrt{ab} = 0$ $(\sqrt{a} - \sqrt{b})^2 = 0$ 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 relation between $a$ and $b$ is:

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