10. If the inequality $x^{2}-\log _{a} x<0$ holds for all $x$ in the interval $\left(0, \frac{1}{2}\right]$, then the range of the real number $a$ is . $\qquad$
10. $\left(\frac{1}{16}, 1\right)$.

When $x \in\left(0, \frac{1}{2}\right]$, $x^{2} \in\left(0, \frac{1}{4}\right]$. Also, on the interval $\left(0, \frac{1}{2}\right]$, $\log _{a} x>x^{2}>0$, then $0$ $\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=\log _{a} a^{\frac{1}{4}}$, we get $a^{\frac{1}{4}}>\frac{1}{2}$, so $a>\frac{1}{16}$. In summary, $\frac{1}{16}<a<1$.