$\begin{aligned} & \left(e^{2 x}-4\right)\left(6 e^{2 x}-3 e^x-2 e^x+1\right)=0 \\ & \left(e^{2 x}-4\right)\left(3 e^x-1\right)\left(2 e^x-1\right)=0 \\ & e^{2 x}=4 \text { or } e^x=\frac{1}{3} \text { or } e^x=\frac{1}{2} \\ & \Rightarrow \text { sum of real roots }=\frac{1}{2} \ln 4+\ln \frac{1}{3}+\ln \frac{1}{2} \\ & =-\ln 3\end{aligned}$