5. Given the function $f(x)=\ln (2+3 x)-\frac{3}{2} x^{2}$, if for any $x \in\left[\frac{1}{6}, \frac{1}{3}\right]$, the inequality $|a-\ln x|+$ $\ln \left[f^{\prime}(x)+3 x\right]>0$ always holds, then the range of the real number $a$ is $\qquad$.
5. $\left\{a \left\lvert\, a \neq \ln \frac{1}{3}\right., a \in \mathbf{R}\right\}$ Detailed explanation: The inequality can be transformed as follows: $|a-\ln x|+\ln \frac{3}{3 x+2}>0$ (1, $\because \frac{1}{6} \leqslant x \leqslant \frac{1}{3}$, $\therefore \frac{3}{3 x+2} \geqslant \frac{3}{3 \cdot \frac{1}{3}+2}=1$, thus we must have $|a-\ln x|+\ln \frac{3}{3 x+2} \geqslant 0$ (2), and it is only when $\left\{\begin{array}{l}x=\frac{1}{3}, \\ a=\ln x,\end{array}\right.$, i.e., $a=\ln \frac{1}{3}$, that $|a-\ln x|+\ln \frac{3}{3 x+2}=0$, Therefore, the range of real number $a$ is $\left\{a \left\lvert\, a \neq \ln \frac{1}{3}\right., a \in \mathbf{R}\right\}$.