Question

The number of real roots of the equation $e^{4x}+{2e}^{3x}-e^x-6=0$ is

• A. $1$
• B. $4$
• C. $2$
• D. $0$

Answer 1

The correct option is A $1$

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

Let $e^x=t$
$\Rightarrow t^4+2t^3-t-6=0$
$\Rightarrow t^4+2t^3+t^2-t^2-t-6=0$
$\Rightarrow (t^2+t{)}^2-(t^2+t)-6=0$

Let $t^2+t=z$
$\Rightarrow z^2-z-6=0$
$\Rightarrow (z-3)(z+2)=0$
$\Rightarrow z=3\text{or}z=-2$
$\Rightarrow t^2+t-3=0\text{or}{t}^2+t+2=0\text{(rejected)}$
$\Rightarrow t=\frac{-1\pm \sqrt{13}}{2}$

Since, $e^x=t>0$
$\therefore e^x=\frac{-1+\sqrt{13}}{2}$

Hence, the given equation has only one real root.

Alternate Solution:
Let $e^x=t$

Now, assuming
$f\left(t\right)=t^4+2t^3-t-6,t>0\Rightarrow f^{\prime }\left(t\right)=4t^3+6t^2-1\Rightarrow f^{\prime \prime }\left(t\right)=12t^2+12t>0[\because t>0]$

So, $f^{\prime }\left(t\right)$ is always increasing.

Also, $f^{\prime }\left(0\right)=-1<0,f^{\prime }\left(1\right)=9>0$

So, $f^{\prime }\left(t\right)=0,$ for only one value of $t\in (0,1).$

Now, the nature of the graph is
$f\left(0\right)=-6<0f\left(1\right)=-4<0f\left(2\right)>0$


Hence, $f\left(t\right)=0$ has only one real solution.