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 $$\log_{x}a + \log_{y}a = 4\log_{xy}a.$$
Umm, I have no idea. I don't think any identities can help here.
Either there is something wrong with the question or I have misunderstood it, because the answer seems to be obviously $0$.
Short justification: put $x=y=\frac12$ and $a=\frac14$. Then $log_\frac12\frac14=2$ and $\log_\frac14\frac14=1$, and your result follows.
Long justification: $\log_x a$ is identical to $\frac{\log a}{\log x}$, where $\log$ represents a logarithm to any base you feel like: $10$ or $e$ or anything else. So your equation becomes $$\frac{\log a}{\log x}+\frac{\log a}{\log y}=4\frac{\log a}{\log x + \log y}\text.$$
Dividing by $\log a$ and rearranging, this gives $$\frac{\log x + \log y}{\log x\log y}=4\frac{1}{\log x + \log y}\text.$$ Multiplying by $(\log x+\log y)\log x\log y$, $$(\log x + \log y)^2=4\log x\log y\text.$$ Multiplying out the left-hand side, $$(\log x)^2 +2\log x \log y + (\log y)^2=4\log x\log y\text.$$ This gives $$(\log x)^2 -2\log x \log y + (\log y)^2=0\text,$$ which gives $$(\log x - \log y)^2=0\text:$$ in other words, $\log x=\log y$.
So all your ordered pairs will have $x$ and $y$ equal. You can see why I feel someone must have missed something here!