2
$\begingroup$

I am really stumped by this equation. $$e^x=x$$ I need to prove that this equation has no real roots. But I have no idea how to start.

If you look at the graphs of $y=e^x$ and $y=x$, you can see that those graphs do not meet anywhere. But I am trying to find an algebraic and rigorous proof. Any help is appreciated. ( $e^\infty=\infty$ looks like a solution but is it?)

$\endgroup$
3
  • $\begingroup$ "Do not meet" is very different from "meets at infinity". Please clarify which one you think is correct - they can't both be true. $\endgroup$
    – Nij
    Oct 26, 2016 at 11:10
  • $\begingroup$ Either you are playing in an extension of $\mathbb R$ where $\infty$ is a point, for topological or algebraical purpose, and you have to mention it, please (you learn everyday here :-), or you just work in standard $\mathbb R$, then please stop saying some nonsense like "one obvious solution would be $x=\infty$", because it's definitely not obvious, and even more definitely not a solution !!! Just study the function $f:x\mapsto e^x-x$. $\endgroup$ Oct 26, 2016 at 11:17
  • $\begingroup$ @Nij Point noted. $\endgroup$
    – S R Maiti
    Nov 2, 2016 at 16:48

5 Answers 5

12
$\begingroup$
  1. For $x \le 0$ we have $e^x >0$, thus no solution of the above equation exists in $(-\infty, 0)$

  2. If $x>0$, then $e^x=1+x+\frac{x^2}{2!}+\ldots>x$. Therefore: no solution of the above equation exists in $(0, \infty)$.

$\endgroup$
5
$\begingroup$

Let $f(x) = e^x - x$. $f'(x) = e^x -1 \geq 0$ and hence $f$ is increasing. Thus for any $x > 0$, $f(x) > f(0)$ and hence $e^x > x+1$ and $f(x) = 0$ has no solution when $x > 0$. For $x <0$, $e^x >0$ and $x < 0$ and hence $e^x \neq x$.

$\endgroup$
2
  • $\begingroup$ $e^{-1}-1<0$. It is decreasing for every $x<0$. $\endgroup$ Oct 26, 2016 at 13:39
  • $\begingroup$ However, the derivative has only one zero, at which the function attains a local minimum of $1$ at $x=0$. $\endgroup$ Oct 26, 2016 at 13:40
5
$\begingroup$

Study $f : x \mapsto e^x - x$

Its derivative is $x \mapsto e^x-1$ which cancels at $0$, is negative before and positive after, thus $f$ reaches a minimum at $0$, and that minimum is $f(0) = 1 \gt 0$. Thus $f$ never cancels, and there are therefore no real solutions to the equation $e^x -x = 0$ ie $e^x = x$

$\endgroup$
1
$\begingroup$

At $x=0$ , $e^x>x$ now derivative of $y=x$ equals $1$ while derivative of $e^x$ is $e^x$ which is increasing and always positive thus the two graphs never meet.

$\endgroup$
1
$\begingroup$

We have $e^x=x$ if and only if $x=\ln x$, which means, for one thing that there can be no solution with $x\le0$. For $x\gt0$, we have

$$\ln x=\int_1^x{dt\over t}\le\int_1^xdt=x-1\lt x$$

so there is no solution for $x\gt0$ either.

Remark: The key inequality between the two integerals may look puzzling for $x\lt1$, since ${1\over t}\gt1$ in that case; but it's still correct because the integrals are both negative.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .