<s>[INST] Unable to justify solution for a problem with exponential constant $e$ involved. For the question,
$e^{2x} + e^x - 2 = 0.$
I was asked to solve for $x$.
What I performed,
$e^{2x} + e^x = 2.$
$e^x(e^x + 1) = 2.$
For the 2 solutions involved,
\begin{align}
e^x &= 2\\
\ln2 &= x\\
.69 &= x
\end{align}
OR 
\begin{align}
e^x + 1 &= 2\\
    e^x &= 1\\
    \ln1 &= x\\
      0 &= x
\end{align}
My textbook states a single solution which is $0$.
 [/INST] Let $y=e^{x}$. Then, your equation becomes
$$
y^{2}+y-2=0\implies (y+2)(y-1)=0.
$$
Thus, $y=-2$ or $y=1$. Hence, $e^{x}=-2$ or $e^{x}=1$. Since the first is impossible, it follows that $e^{x}=1\implies x=0$. 
</s>