The equation e ^ 4x + 8 e ^ 3x + 13 e ^ 2x - 8 e ^ x + 1 = 0, x ∈ R...

Given,

${e}^{4x} + 8{e}^{3x} + 13{e}^{2x} - 8{e}^{x} + 1 = 0, x \in R$

Now let ${e}^{x} = t$ we get,

${t}^{4} + 8{t}^{3} + 13{t}^{2} - 8t + 1 = 0$

Now divide complete equation by ${t}^{2}$

$⇒ {t}^{2} + \dfrac{1}{{t}^{2}} + 8\left(t - \dfrac{1}{t}\right) + 13 = 0$

$⇒ {\left(t - \dfrac{1}{t}\right)}^{2} + 8\left(t - \dfrac{1}{t}\right) + 15 = 0$

Now let $t - \dfrac{1}{t} = z$ we get,

$⇒ {z}^{2} + 8z + 15 = 0$

$⇒ z = - 3, - 5$

So, $t - \dfrac{1}{t} = - 3$ or $t - \dfrac{1}{t} = - 5$

$⇒ {t}^{2} + 3t - 1 = 0$ or ${t}^{2} + 5t - 1 = 0$

$⇒ t = \dfrac{- 3 \pm \sqrt{13}}{2}, \dfrac{- 5 \pm \sqrt{29}}{2}$

So, ${e}^{x} = \dfrac{- 3 + \sqrt{13}}{2}, \dfrac{- 5 + \sqrt{29}}{2} = \alpha, \beta$ (rejecting negative values as exponential is positive function)

And both $\dfrac{- 3 + \sqrt{13}}{2}$ and $\dfrac{- 5 + \sqrt{29}}{2} \in \left(0, 1\right)$

So, $x = \ln \left(\alpha\right), \ln \left(\beta\right)$ are both negative,

Hence, there are two solution and both are negative.