The sum of all real roots of equation ( e ^ 2x - 4)(6 e ^ 2x - 5 e...

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

So, the given equation becomes, \left({t}^{2} - 4\right)\left(6{t}^{2} - 5t + 1\right) = 0$</p><p>${t}^{2} - 4 = 0$(or)$6{t}^{2} - 5t + 1 = 0$</p><p>${t}^{2} = 4$(or)$6{t}^{2} - 3t - 2t + 1 = 0$</p><p>$t = \pm 2$(or)$3t\left(2t - 1\right) - 1\left(2t - 1\right) = 0$</p><p>$t = \pm 2$(or) \left(3t - 1\right)\left(2t - 1\right) = 0$

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

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

$∴$ Only possible values for $t$ are $2, \dfrac{1}{3}, \dfrac{1}{2}$

When $t = 2, {e}^{x} = 2 ⇒ x = \log_{e}\left(2\right)$ (or) $\ln 2$

When $t = \dfrac{1}{3}, {e}^{x} = \dfrac{1}{3} ⇒ x = \log_{e}\left(\dfrac{1}{3}\right)$ or $- \ln 3$

When $t = \dfrac{1}{2}, {e}^{x} = \dfrac{1}{2} ⇒ x = \log_{e}\left(\dfrac{1}{2}\right)$ or $- \ln 2$

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