Find the minimum of $x-y$ among all ordered pairs of real numbers $(x, y)$, $x$ and $y$ between 0 and 1, where there exists a real number $a \neq 1$ such that $\log_{x}a + \log_{y}a = 4\log_{xy}a.$
If someone could please answer the above question, I would greatly appreciate it.
Given $$ \log _{x}a + \log _{y}a = 4 \log _{xy}a \\ \frac{\ln a}{\ln x} + \frac{\ln a}{\ln y} = \frac{4 \ln a}{\ln xy} \\ \frac{\ln xy}{\ln x} + \frac{\ln xy}{\ln y} = 4 \qquad a \neq e \\ \frac{\ln x + \ln y}{\ln x} + \frac{\ln x + \ln y}{\ln y} = 4 \\ \frac{\ln x}{\ln y} + \frac{\ln y}{\ln x} = 2 \\ (\ln x)^2 + (\ln y)^2 = 2 \ln x \ln y \\ (\ln x - \ln y)^2 = 0 \\ x = y \\ \text{then it follows that} \quad 0 \quad \text{is the minimum (and maximum) of} \quad x - y \quad \text{for} \quad x,y \in (0,1). $$