7.116. $1+2 \log _{x} 2 \cdot \log _{4}(10-x)=\frac{2}{\log _{4} x}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ x<10 .\end{array}\right.$

By changing to base 2, we have

$1+\frac{\log _{2}(10-x)}{\log _{2} x}=\frac{4}{\log _{2} x}, \log _{2} x+\log _{2}(10-x)=4, \quad \log _{2} x(10-x)=4 \Rightarrow$

$\Rightarrow x^{2}-10 x+16=0$,

from which $x_{1}=2, x_{2}=8$.

Answer: $2 ; 8$.