10. The number of integer solutions to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$ is $\qquad$ .
10. 24 .

Let $\log _{25} x=t$. Then $x=25^{t}$.
Thus the original inequality $\Leftrightarrow \log _{6}\left(1+5^{t}\right)>t$
$$
\Leftrightarrow 1+5^{t}>6^{t} \Leftrightarrow\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1 \text {. }
$$

Define $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}$. Then $f(t)$ is a decreasing function, $f(t)>1=f(1)$.
Hence $t=\log _{25} x<1$.
Therefore, $0<x<25$, a total of 24.