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

Let $e^x=t$

Now, $t^4+8 t^5+13 t^2-8 t+1=0$

Dividing equation by $t ^2$,

$t^2+8 t+13-\frac{8}{t}+\frac{1}{t^2}=0$

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

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

Let $t-\frac{1}{t}=z$

$z^2+8 z+15=0$

$(z+3)(z+5)=0$

$z=-3 \text { or } z=-5$

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

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

$t=\frac{-3 \pm \sqrt{13}}{2} \text { or } t =\frac{-5 \pm \sqrt{29}}{2}$

as $t = e ^{ x }$ so $t$ must be positive,

$t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}$

So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$

Hence two solution and both are negative.