What is the solution of x*exp(-x)=1?

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Homework Help Overview

The discussion revolves around solving the equation x*exp(-x)=1, which involves exponential functions and logarithmic properties.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss taking the natural logarithm of both sides and simplifying the equation. There is a suggestion to consider whether a solution exists or if it may be necessary to demonstrate that no solution is possible.

Discussion Status

Participants are exploring different interpretations of the problem, including the possibility of showing that no solution exists. Some guidance has been offered regarding analyzing the function f(x)=ln(x)-x and its derivatives to investigate potential solutions.

Contextual Notes

There is a mention of constraints regarding the domain of x, specifically that x>0 is the only possibility being considered.

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Homework Statement


What is the solution of the equation x*exp(-x)=1?


Homework Equations





The Attempt at a Solution


I tried taking the ln of both sides...and I got
ln(x*exp(-x))=ln(1)
ln(x)-ln(exp(x))=ln(1)
ln(x)-x=ln(1)

Then I don't know what to do next...am I on the right track at least?

Thanks for the help! :)
 
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[tex]xe^{-x}=1[/tex]

Use what you know about negative exponents to rewrite this problem in a more simple form. Where is this problem from? You may need to show that there is no solution rather than finding the solution.
 
futurebird is right. You may have to show it doesn't have a solution. Clearly x>0 is the only possibility. So you got ln(x)-x=ln(1)=0. Take f(x)=ln(x)-x. Look at f'(x) to figure out where it has a max/min etc. Can it ever be zero?
 
x=1 works.

Ooh- my bad- misread the question. It doesn't have a solution.
 
christianjb said:
x=1 works.

You scared me for a moment there!


An easy way to see that there is so solution is to first simplify then graph both sides of the equation by setting them equal to y.
 
Graphing them both and showing the don't intersect is not really the most rigorous proof...

Use Dick's method in post 3.
 
ahhhh...thanks guys! Why didnt I think of that? =\
 

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