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 $0$.
 A: 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 $e^{x}=1\implies x=0$. 
A: $e^x (e^x+1) = 2$ does not imply that $e^x = 2$ or $e^x+1 = 2$.  This type of reasoning only works if the right-hand side is zero.  (This is why it's called the zero-product property.)
In general, if $ab = 2$ then we could have $a=2$ and $b=1$, or $a=1$ and $b=2$, or $a=4$ and $b=1/2$, or $a=2\pi$ and $b=\frac1\pi$, etc.  There are infinitely many possibilities if the right-hand side is not zero, which is why we need to use a different method in such cases.
The proper way to proceed with $e^{2x} + e^x - 2 = 0$ is to first rewrite $e^{2x}$ as $(e^x)^2$.  Then the equation becomes
$$ (e^x)^2 + e^x - 2 = 0.$$
Now we can make the substitution $y = e^x$ to get
$$ y^2 + y - 2 = 0.$$
So now it's a plain quadratic equation.  Factor to get $(y+2)(y-1) = 0$.  Now we can use the zero-product property to say $y = -2$ or $y=1$.  Back-substitute to get
$$ e^x = -2 \qquad \text{ or } \qquad e^x = 1.$$
Since $e^x$ can't be negative for any real value of $x$, the only valid case is $e^x = 1$, and this happens when $x = 0$.
A: It should be $${ e }^{ 2x }+{ e }^{ x }-2=0\\ \left( { e }^{ x }+2 \right) \left( { e }^{ x }-1 \right) =0\\ { e }^{ x }+2=0,{ e }^{ x }-1=0\\ { e }^{ x }=-2,{ e }^{ x }=1$$
is only solution

$$x=0$$

A: From this step:
$e^x(e^x + 1) = 2$,
your solution makes no sense.
Observe that putting $y = e^x$ allows you to transform the original equation to the quadratic $y^2 + y -2 = 0$. This has roots $1$ and $-2$. Only the positive root allows a real solution for $x$, i.e. $e^x = 1$ giving $x=0$ as the only real solution.
