7
$\begingroup$

How can I solve this equation:

$$\ln x+x=1$$

We had it on a local Olympiad math contest problem.

$\endgroup$
13
$\begingroup$

$1$ 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

$\endgroup$
  • $\begingroup$ thanks, did you ever have math contest training? $\endgroup$ – user144251 Apr 19 '14 at 14:53
  • 1
    $\begingroup$ @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
5
$\begingroup$

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 $x=1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @user144251 "That was so obvious!!". Is that a compliment or what? $\endgroup$ – Anastasiya-Romanova 秀 Apr 19 '14 at 15:05
4
$\begingroup$

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

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 $x=1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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