Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
$\ln(x+e^{2x}) \neq \ln x + \ln e^{2x}$. –  Javier Badia Jun 2 '13 at 16:05
add comment

4 Answers

up vote 3 down vote accepted

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

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

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

$\Rightarrow f$ 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 $x=0$.

share|improve this answer
add comment

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 $y = 1$)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.