Express as a single logarithm and simplify, if possible.
$$\frac{1}{4}{\log }_{e}x+3{\log }_{e}y-4{\log }_{e}x$$

Step 1 ：Given the expression $\frac{1}{4}{\log }_{e}x+3{\log }_{e}y-4{\log }_{e}x$, we can use the properties of logarithms to simplify.

Step 2 ：First, we can combine the terms involving ${\log }_{e}x$: $\frac{1}{4}{\log }_{e}x-4{\log }_{e}x=-\frac{15}{4}{\log }_{e}x$.

Step 3 ：So the expression becomes $-\frac{15}{4}{\log }_{e}x+3{\log }_{e}y$.

Step 4 ：Next, we can use the power rule of logarithms, which states that $a{\log }_{b}c={\log }_b(c^a)$, to rewrite the expression as ${\log }_e{x}^{-\frac{15}{4}}+{\log }_e{y}^3$.

Step 5 ：Finally, we can use the product rule of logarithms, which states that ${\log }_{b}m+{\log }_{b}n={\log }_b(mn)$, to combine the two terms into a single logarithm: ${\log }_e(x^{-\frac{15}{4}}{y}^3)$.

Step 6 ：Thus, the expression $\frac{1}{4}{\log }_{e}x+3{\log }_{e}y-4{\log }_{e}x$ simplifies to $\boxed{{\displaystyle {\log }_e(x^{-\frac{15}{4}}{y}^3)}}$.