12. Find the real solution to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$,
12. Let $t=\log _{25} x$, then $x=25^{t}$, substituting into the original equation gives $\log _{6}\left(1+5^{t}\right)>t$, which means $1+5^{t}>6^{t}$, or $\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1$. Let $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5^{t}}{6}\right)$, $f(t)$ is a decreasing function, and $f(t)>1=f(1)$, so $t<1$, i.e., $\log _{25} x<1,0<x<25$.