Answer: 25

Small Hint:

Bring the exponents down: $x \log_x y = 10$ and $4y \log_y x = 10$

Big Hint:

Multiply the two equations: since $\log_x y \cdot \log_y x = 1,$ the logarithms disappear

Solution:

Pulling the exponents out of the logarithms, the conditions become $$x \log_x y = 10$$ and $$4y \log_y x = 10.$$ Multiplying these equations and using $\log_x y \cdot \log_y x = 1$ gives $4xy = 100,$ so $xy = 25.$

Such $x$ and $y$ do exist. Set $y = \frac{25}{x}.$ The continuous function $x\log_x(\frac{25}{x})$ tends to infinity as $x \to 1^+,$ while at $x = 2$ it is less than $10.$ Hence it equals $10$ for some $1 \lt x \lt 2,$ where $y \gt 1.$