# 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!

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$\ln(x+e^{2x}) \neq \ln x + \ln e^{2x}$. – Javier Jun 2 '13 at 16:05

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.

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Ok got it, if $f'(x)>0$ is only increasing and if $f(a)*f(b)<0$ so there is $f(c)=0$ – Ofir Attia Jun 2 '13 at 10:09
@OfirAttia Yes. – Git Gud Jun 2 '13 at 10:10

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.

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$f(x)=x+e^{2x}-1$

$f'(x)=1+2e^{2x}>0 \forall x\in R$

$\Rightarrow f$ 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 $x=0$.

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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 $y = 1$)

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