Is this equation possible to solve exactly?

  • Thread starter Thread starter Hertz
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the equation ln(x) = x - 2, focusing on the possibility of solving it exactly. Participants explore the nature of solutions, particularly in relation to irrational numbers and the use of special functions like the Lambert W function.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants question the meaning of "solving exactly" and discuss the implications of irrational solutions. Some mention the use of the product log function and the Lambert W function as potential methods for expressing solutions. Others suggest numerical methods like Newton's method for approximating solutions.

Discussion Status

The discussion is active, with various interpretations of what constitutes an exact solution being explored. Some participants provide insights into the use of special functions, while others express curiosity about the analytical methods available for such equations.

Contextual Notes

There is an ongoing debate about the nature of exact solutions versus numerical approximations, with references to known irrational numbers and their representation. The discussion also touches on the limitations of expressing certain solutions analytically.

Hertz
Messages
180
Reaction score
8
[itex]ln(x) = x - 2[/itex]

Any ideas?
 
Physics news on Phys.org
I plugged it into wolfram alpha and it gave me two solutions in the form of values in the product log function: a function I had never heard of but seems to not be solvable exactly.
 
What do you mean by "solve exactly"? It has two well-defined solutions, and you could calculate their values to as many decimal places as you want, but they are irrational numbers, so you could never write them down exactly, just like you could never write down pi or sqrt(2) exactly. Is it possible to solve x^2 = 2 exactly?
 
phyzguy said:
What do you mean by "solve exactly"? It has two well-defined solutions, and you could calculate their values to as many decimal places as you want, but they are irrational numbers, so you could never write them down exactly, just like you could never write down pi or sqrt(2) exactly. Is it possible to solve x^2 = 2 exactly?

You can write down pi and sqrt(2) exactly with no problems whatsoever. Here, I'll show you:

pi
sqrt(2)

However, if you attempted to solve an equation who's answer was pi, but you had no way to analytically prove that it was pi, you would be unable to give an exact answer to the problem.

I want to know if there is an analytic method to solve an equation like this, just like there are analytic methods to exactly solve many other equations with irrational answers.

Back to the original question, do you have any ideas?

If you think the answer to this problem truly is irrational and simply cannot be evaluated exactly (with the use of e, pi, sqrts, and other such symbols/constants), then please let me know the reasoning behind your hypothesis.

e-
Let me reword my question.
Is it possible to analytically solve for x in the equation above? If so, how could it be done?
 
Hertz said:
You can write down pi and sqrt(2) exactly with no problems whatsoever. Here, I'll show you:

pi
sqrt(2)

I can also write down the solution to your equation. Here, I'll show you:

-ProductLog(-1/e^2)

Does that answer your question?
 
phyzguy said:
I can also write down the solution to your equation. Here, I'll show you:

-ProductLog(-1/e^2)

Does that answer your question?

Oh thank you so much :) I'm happy I finally know how to analytically solve an equation like this when I come across one. The method you taught me is so simple.
 
You could use Newton's method to find an approximation.
 
TheEtherWind said:
You could use Newton's method to find an approximation.

You know that's the first time I've ever seen a good legitimate use for Newton's method. I guess there was a reason I learned it after all.
 
Hertz said:
[itex]ln(x) = x - 2[/itex]

Any ideas?

You can solve it in terms of the Lambert W function. Here is what Maple gives:

Solution 1:
x = exp(-LambertW(-exp(-2))-2)

Solution 2:
x = exp(-LambertW(-1,-exp(-2))-2).

RGV
 
  • #10
The "Lambert W function" is another name for the "product log" that phyzguy mentioned.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K