$8.58 \log _{4} x+\log _{2}(\sqrt{x}-1)<\log _{2} \log _{\sqrt{5}} 5$ (find integer values of $x$).
8.58 Given the domain of the logarithmic function and the square root, we conclude that $x>0$ and $\sqrt{x}-1>0$, hence $x>1$. We will convert all logarithms to base 2: $\log _{2} \sqrt{x}+\log _{2}(\sqrt{x}-1)0$ and solve the inequality $y^{2}-y-21$ and, therefore, $1<\sqrt{x}<2$, i.e., $1<x<4$. The integer values of $x$ that satisfy this inequality are 2 and 3.

Answer: 2; 3.