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

Let ${e}^{x}=t$ex=t so, ${e}^{2x}={t}^{2}$e2x=t2

So, the given equation becomes, $({t}^{2}-4)(6{t}^{2}-5t+1)=0$(t2−4)(6t2−5t+1)=0

${t}^{2}-4=0$t2−4=0 (or) $6{t}^{2}-5t+1=0$6t2−5t+1=0

${t}^{2}=4$t2=4 (or) $6{t}^{2}-3t-2t+1=0$6t2−3t−2t+1=0

$t=\pm 2$t=±2 (or) $3t(2t-1)-1(2t-1)=0$3t(2t−1)−1(2t−1)=0

$t=\pm 2$t=±2 (or) $(3t-1)(2t-1)=0$(3t−1)(2t−1)=0

$t=\pm 2$t=±2 (or) $t=\frac{1}{3}$t=31​ (or) $t=\frac{1}{2}$t=21​

But, ${e}^{x}=t>0(\because {e}^{x}>0;\forall x\in R)$ex=t>0(∵ex>0;∀x∈R)

$\therefore$∴ Only possible values for $t$t are $2,\frac{1}{3},\frac{1}{2}$2,31​,21​

When $t=2,{e}^{x}=2\Rightarrow x={\mathrm{log}}_{e}(2)$t=2,ex=2⇒x=loge​(2) (or) $\mathrm{ln}2$ln2

When $t=\frac{1}{3},{e}^{x}=\frac{1}{3}\Rightarrow x={\mathrm{log}}_{e}(\frac{1}{3})$t=31​,ex=31​⇒x=loge​(31​) or $-\mathrm{ln}3$−ln3

When $t=\frac{1}{2},{e}^{x}=\frac{1}{2}\Rightarrow x={\mathrm{log}}_{e}(\frac{1}{2})$t=21​,ex=21​⇒x=loge​(21​) or $-\mathrm{ln}2$−ln2

$\therefore$∴ The sum of roots of given equation $=\mathrm{ln}2-\mathrm{ln}3-\mathrm{ln}2=-\mathrm{ln}3$=ln2−ln3−ln2=−ln3.