10. Given the function
$$
f(x)=\left\{\begin{array}{ll}
(3-m) x-m, & x<1 ; \\
\log _{m} x, & x \geqslant 1
\end{array}\right.
$$

is monotonically increasing on the real number set $\mathbf{R}$. Then the range of the real number $m$ is.
10. $\frac{3}{2} \leqslant m1, \\
3-m>0, \\
\log _{m} 1 \geqslant(3-m) \times 1-m .
\end{array}\right.
$

Solving gives $\frac{3}{2} \leqslant m<3$.