Prove that $x+e^{2x}=1$ have only one solution I`m trying to prove that this equation have only one solution.
$$x+e^{2x}=1$$
so what I did is to set $\ln$ on this equation and get:
$$\ln(x)+2x=0$$
I need some hint how to continue from here.
Thanks!
 A: Hint: Ignore what you did. Consider the function $f(x)=x+e^{2x}-1$. Relate this function to your problem somehow and use the intermediate value theorem. This takes care of the existence of one solution. To ensure it's unique, think about $f'$.

Regarding your work, note that the equations you got aren't equivalent due to the fact that the LHS of the first equation makes sense on a bigger set than the LHS of the second equation.
A: Hints:
Define
$$f(x):=x+e^{2x}-1\implies f'(x)=1+2e^{2x}>0\,\,\forall\,x\in\Bbb R\;\implies$$
the function is monotone ascending and thus has at most one zero...which it obviously has.
A: $f(x)=x+e^{2x}-1$
$f'(x)=1+2e^{2x}>0 \forall x\in R$
$\Rightarrow f$ is an increasing function and as $f(0)=0$
So $\forall x,y\in R$ with $x<0<y$ we must have $f(x)<0<f(y)$
Hence there is only one solution namely $x=0$. 
A: Another way is to notice that your problem can be regarded as system
$$
\left\{
\begin{array}{rcl}
y &=& e^{2x}\\
y &=& 1 - x \\
\end{array}\right.
$$ 
Notice that the first function is increasing and the second is decreasing. Also from plots of these functions you will see that the system has only one solution ($x = 0$ and $y = 1$) 
