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

Given,

$${e}^{4x}+8{e}^{3x}+13{e}^{2x}-8{e}^{x}+1=0,x\in R$$e4x+8e3x+13e2x−8ex+1=0,x∈R

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

$${t}^{4}+8{t}^{3}+13{t}^{2}-8t+1=0$$t4+8t3+13t2−8t+1=0

Now divide complete equation by ${t}^{2}$t2

$$\Rightarrow {t}^{2}+\frac{1}{{t}^{2}}+8(t-\frac{1}{t})+13=0$$⇒t2+t21​+8(t−t1​)+13=0

$$\Rightarrow {(t-\frac{1}{t})}^{2}+8(t-\frac{1}{t})+15=0$$⇒(t−t1​)2+8(t−t1​)+15=0

Now let $t-\frac{1}{t}=z$t−t1​=z we get,

$$\Rightarrow {z}^{2}+8z+15=0$$⇒z2+8z+15=0

$$\Rightarrow z=-3,-5$$⇒z=−3,−5

So, $t-\frac{1}{t}=-3$t−t1​=−3 or $t-\frac{1}{t}=-5$t−t1​=−5

$\Rightarrow {t}^{2}+3t-1=0$⇒t2+3t−1=0 or ${t}^{2}+5t-1=0$t2+5t−1=0

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

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

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

So, $x=\mathrm{ln}(\alpha ),\mathrm{ln}(\beta )$x=ln(α),ln(β) are both negative,

Hence, there are two solution and both are negative.