exponential equation with a sum of exponents I'm trying to solve the following exponential equation:
$e^{2x} - e^{x+3} - e^{x + 1} + e^4 = 0$
According to the the text I am using the answer should be $x = 1,3$ but I can't derive the appropriate quadratic $x^2 -4x + 3$ from the above equation using any of the methods I know. Can someone point me in the right direction? Is there a substitution I'm not seeing?
 A: $$e^{2x} - e^{x+3} - e^{x + 1} + e^4 = 0$$
$$\left(e^{x+1}\right)\left(e^{2x-\left(x+1\right)}-e^{\left(x+3\right)-\left(x+1\right)}-e^{\left(x+1\right)-\left(x+1\right)}+e^{4-\left(x+1\right)}\right)=0$$
$$\left(e^{x+1}\right)\left(e^{x-1}-e^{2}-e^{0}+e^{3-x}\right)=0$$
$$\left(e^{x+1}\right)\left(e^{x-1}-e^{2}-1+e^{3-x}\right)=0$$
$$\left(e^{x+1}\right)\left(e^{-x-1}\right)\left(e^x-e\right)\left(e^x-e^3\right)=0$$
$$e^0 \cdot \left(e^x-e\right)\left(e^x-e^3\right) =0$$
$$\left(e^x-e\right)\left(e^x-e^3\right)=0$$
Thus, either $e^x-e=0$ or $e^x-e^3 = 0$. These correspond to $x=1$ or $x=3$.
A: $$e^{2x} - (e^3 +e)e^x + e^4$$
Let $m = e^x$
$$m^2- (e^3 + e)m + e^4$$
$a = 1$, $b=-(e^3 + e)$, $c =e^4$
$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{(e^3 + e) \pm \sqrt{{(-(e^3 + e))}^2 -4(1)(e^4)}}{2(1)}$$ 
$$= \frac{(e^3 + e) \pm \sqrt{e^6 +e^2 + 2e^4 - 4e^4}}{2}$$ 
$$= \frac{(e^3 + e) \pm \sqrt{e^4 +1 + 2e^2 - 4e^2} \sqrt{e^2}} {2}$$ 
$$= \frac{(e^3 + e) \pm \sqrt{e^4 - 2e^2 + 1} \sqrt{e^2}} {2}$$ 
$$= \frac{(e^3 + e) \pm \sqrt{(e^2 - 1)^2} \cdot e} {2}$$
$$=\frac{e^3 + e \pm e^2 - 1 \cdot e}{2}$$
$$=\frac{e^3 + e \pm (e^3 -e)}{2}$$
$$m = \frac{e^3 +e + e^3 - e}{2}, \frac{e^3 + e - e^3 + e}{2}$$
$$\implies m = \frac{2e^3}{2}, \frac{2e}{2}$$
$$m = e^3, e$$
But $m= e^x$
$$e^x = e^3, e^x =e$$
Since the bases are the same, you can equate $x = 1,3$
