19. Given the function $f(x)$ for any $s, t \in \mathbf{R}$, it satisfies $f(s+t)-f(t)=(s+2 t+1) s$, and $f(1)=0$. Also, when $x \in\left(0, \frac{1}{2}\right)$, the inequality $f(x)<\log _{a} x-2$ always holds. Try to find the range of the real number $a$.
19. Let $s=1, t=0$, then $f(1)-f(0)=2$, combining with $f(1)=0$ we get $f(0)=-2$,

then $f(x)+2=f(x)-f(0)=f(x+0)-f(0)=(x+0+1) x=(x+1) x$, so when $x \in\left(0, \frac{1}{2}\right)$ the inequality $(x+1) x<\log _{2} x$ always holds, combining the graphical method we get
$$
\left\{\begin{array}{l}
0<a<1, \\
\left(\frac{1}{2}+1\right) \frac{1}{2} \leqslant \log _{a} \frac{1}{2},
\end{array}\right.
$$

solving for the range of real number $a$ we get
$$
\left\{a \left\lvert\, \frac{\sqrt[3]{4}}{4} \leqslant a<1\right.\right\} .
$$