Let ex=t so, e2x=t2
So, the given equation becomes, (t2−4)(6t2−5t+1)=0
t2−4=0 (or) 6t2−5t+1=0
t2=4 (or) 6t2−3t−2t+1=0
t=±2 (or) 3t(2t−1)−1(2t−1)=0
t=±2 (or) (3t−1)(2t−1)=0
t=±2 (or) t=31 (or) t=21
But, ex=t>0(∵ex>0;∀x∈R)
∴ Only possible values for t are 2,31,21
When t=2,ex=2⇒x=loge(2) (or) ln2
When t=31,ex=31⇒x=loge(31) or −ln3
When t=21,ex=21⇒x=loge(21) or −ln2
∴ The sum of roots of given equation =ln2−ln3−ln2=−ln3.