The sum of all real roots of equation e 2 x - 4 6 e 2 x - 5 e x + 1 = 0 is

Let $e^x=t$  so, $e^{2x}=t^2$

So, the given equation becomes, $\left(t^2-4\right)\left(6t^2-5t+1\right)=0$

$t^2-4=0$ (or) $6t^2-5t+1=0$

$t^2=4$   (or) $6t^2-3t-2t+1=0$

$t=\pm 2$ (or) $3t\left(2 t-1\right)-1\left(2 t-1\right)=0$

$t=\pm 2$ (or)  $\left(3t-1\right)\left(2t-1\right)=0$

$t=\pm 2$ (or) $t=\frac{1}{3}$ (or) $t=\frac{1}{2}$

But, $e^x=t>0 \left(\because e^x>0;\forall x\in R\right)$

$\therefore$  Only possible values for $t$ are $2,\frac{1}{3},\frac{1}{2}$

When $t=2, e^x=2 \Rightarrow x={\log }_e\left(2\right)$ (or) $\ln 2$

When $t=\frac{1}{3}, e^x=\frac{1}{3} \Rightarrow x={\log }_e\left(\frac{1}{3}\right)$ or $-\ln 3$

When $t=\frac{1}{2}, e^x=\frac{1}{2}\Rightarrow x={\log }_e\left(\frac{1}{2}\right)$ or $-\ln 2$

$\therefore$  The sum of roots of given equation $=\ln 2-\ln 3-\ln 2=-\ln 3$.