7.189. $\log _{x} m \cdot \log _{\sqrt{m}} \frac{m}{\sqrt{2 m-x}}=1$.
Solution.

Domain of definition: $\left\{\begin{array}{l}0<m \neq 1, \\ 0<x \neq 1, \\ x<2 m .\end{array}\right.$

Switch to base $m$, then

$\frac{1}{\log _{m} x} \cdot \frac{\log _{m} \frac{m}{\sqrt{2 m-x}}}{\log _{m} \sqrt{m}}=1 \Leftrightarrow \log _{m} x+\log _{m}(2 m-x)=2 \Rightarrow$

$\Rightarrow \log _{m} x(2 m-x)=2$.

Then $x^{2}-2 m x+m^{2}=0,(x-m)^{2}=0$, from which $x=m$.

Answer: $\quad m$, where $0<m \neq 1$.