If log (x + z) + log (x + z - 2y) = 2log (x - | vedclass

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(A) Given: $\log (x + z) + \log (x + z - 2y) = 2\log (x - z)$Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:$\log ((x +

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$H.P.$

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(A) Given: $\log (x + z) + \log (x + z - 2y) = 2\log (x - z)$Using the property $\log a + \log b = \log (ab)$ and $n \log a = \log (a^n)$:$\log ((x + z)(x + z - 2y)) = \log ((x - z)^2)$$(x + z)(x + z - 2y) = (x - z)^2$$(x + z)^2 - 2y(x + z) = x^2 - 2xz + z^2$$x^2 + 2xz + z^2 - 2xy - 2yz = x^2 - 2xz + z^2$$2xz - 2xy - 2yz = -2xz$$4xz = 2xy + 2yz$$2xz = xy + yz$Dividing both sides by $xyz$:$\frac{2xz}{xyz} = \frac{xy}{xyz} + \frac{yz}{xyz}$$\frac{2}{y} = \frac{1}{z} + \frac{1}{x}$This implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$Therefore,$x, y, z$ are in $H.P.$

If $\log (x + z) + \log (x + z - 2y) = 2\log (x - z),$ then $x, y, z$ are in

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