The number of real roots of the equation e 4 x + 2 e 3 x - e x - 6 = 0 is :

Given that

$e^{4x}+2e^{3x}-e^x-6=0$

Let $e^x=t$,  $t>0$

$f\left(t\right)=t^4+2t^3-t-6$
$f\text{'}\left(t\right)=4t^3+6t^2-1$

$f^{\text{'}\text{'}}\left(t\right)=12t^2+12t>0$ $(t>0)$

$f\text{'}\left(t\right)$ is increasing function

Now $f^{\text{'}}\left(0\right)=-1$

$f^{\text{'}}\left(1\right)=9$

By using Intermediate value theorem there exists a root for $f\left(x\right) = 0$if $f\left(\alpha \right)f\left(\beta \right)<0 \Rightarrow f\left(\gamma \right) = 0 , \gamma \in \left(\alpha , \beta \right)$

So $f^{\text{'}}\left(\beta \right)=0$

Hence, only one real root.