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

Given,

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

Now let $e^x=t$ we get,

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

Now divide complete equation by $t^2$

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

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

Now let $t-\frac{1}{t}=z$ we get,

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

$\Rightarrow z=-3,-5$

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

$\Rightarrow t^2+3t-1=0$  or   $t^2+5t-1=0$

$\Rightarrow t=\frac{-3\pm \sqrt{13}}{2},\frac{-5\pm \sqrt{29}}{2}$

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

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

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

Hence, there are two solution and both are negative.