If $\text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y)$, prove that x = y.

Given,

$\Rightarrow \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space x + \text{log} \space y) \\[1em] \Rightarrow \text{log} \space \dfrac{x + y}{2} = \dfrac{1}{2}(\text{log} \space xy) \\[1em] \Rightarrow \text{log} \space \dfrac{x + y}{2} = \text{log} \space (xy)^{\dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{x + y}{2} = (xy)^{\dfrac{1}{2}} \\[1em] \Rightarrow x + y = 2(xy)^{\dfrac{1}{2}} \\[1em]$

Squaring both sides we get,

$\Rightarrow (x + y)^2 = 4(xy) \\[1em] \Rightarrow (x^2 + y^2 + 2xy) = 4xy \\[1em] \Rightarrow x^2 + y^2 + 2xy - 4xy = 0 \\[1em] \Rightarrow x^2 + y^2 - 2xy = 0 \\[1em] \Rightarrow (x - y)^2 = 0 \\[1em] \Rightarrow x - y = 0 \\[1em] \Rightarrow x = y.$

Hence, proved that x = y.