4. Given that the function $f(x)$ is defined on $\mathbf{R}$ as an even function, and for any real number $x$, $f(x+1)=f(1-x)$ holds. When $1 \leqslant x \leqslant 2$, $f(x)=\ln x$. If the equation $f(x)+a x-1=0$ has two distinct real roots for $x \in[3,5]$, then the range of values for $a$ is
4. $\left(\frac{1-\ln 2}{4}, \frac{1}{5}\right]$

As shown in Figure 3, the graphs of the functions $y=f(x)$ and $y=-a x+1$ are drawn, where $P(0,1), G\left(\frac{1}{a}, 0\right)$.
From the graph, we know that when
$$
\left\{\begin{array}{l}
-4 a+1<\ln 2, \\
-5 a+1 \geqslant 0
\end{array} \text {, i.e., } \frac{1-\ln 2}{4}<a \leqslant \frac{1}{5}\right.
$$

the two function graphs have two different intersection points on $x \in[3,5]$.
Therefore, the range of $a$ is $\left(\frac{1-\ln 2}{4}, \frac{1}{5}\right)$.