How can I solve this equation:

$$\ln x+x=1$$

We had it on a local Olympiad math contest problem.


$1$ is a solution, just substitute and check.

$x+\log(x)$ is strictly increasing, hence 1 to 1. Thus, $x=1$ is the only solution for the equation

  • $\begingroup$ thanks, did you ever have math contest training? $\endgroup$ – user144251 Apr 19 '14 at 14:53
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    $\begingroup$ @user144251 No, and I actually consider myself not good at all in hard math contests such as IMO $\endgroup$ – Amr Apr 19 '14 at 14:59
  • $\begingroup$ @user144251 See the book " The Art and Craft of Problem solving" $\endgroup$ – Amr Apr 19 '14 at 15:03
  • $\begingroup$ Thanks, I will definitely check it! $\endgroup$ – user144251 Apr 19 '14 at 15:03

Intuitively \begin{align} \ln x+x&=1\\ \ln x+x\ln e&=1\\ \ln x+\ln e^x&=1\\ \ln (xe^x)&=1\\ xe^x&=e^1\\ xe^x&=1\cdot e^1 \end{align} By comparing LHS and RHS, we will obtain $x=1$.

  • $\begingroup$ That was so obvious!! DId you had a formal maths contest training? $\endgroup$ – user144251 Apr 19 '14 at 14:58
  • $\begingroup$ @user144251 No! Why did you ask that? $\endgroup$ – Anastasiya-Romanova 秀 Apr 19 '14 at 14:59
  • $\begingroup$ Because I want to participate in an IMO and I want some advice! $\endgroup$ – user144251 Apr 19 '14 at 15:00
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    $\begingroup$ @user144251 "That was so obvious!!". Is that a compliment or what? $\endgroup$ – Anastasiya-Romanova 秀 Apr 19 '14 at 15:05

$$\ln x+x=1\implies xe^x = e$$

In other words we need to find a root of $f(x)=xe^x - e$

This function is increasing for $x\gt-1$ thus will have at most one solution in $(-1,\infty)$.

Also since $e^x\gt0$ we must have $x\gt0$.

Now it is easy to see that the one root we need is $x=1$


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