What is the solution of x^2=ln(x)

  • Thread starter karpmage
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In summary, there is no real solution for the equation x^2 = ln(x). This can be proven through an argument and by graphing the two sides of the equation. However, if looking for a complex solution, one can define ln x as the principal value and use a numerical solver to find possible real solutions for x. WolframAlpha can also provide an answer for this type of equation.
  • #1
karpmage
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I was just wondering whether anyone could tell me what's the solution of x^2=ln(x).

This isn't a homework question, I need to know the solution of this equation however for some of the maths that I do in my free time.

Also, I'd be interested to see how the solution is obtained.

Thanks.
 
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  • #2
Hey karpmage.

In this case there is no solution.

Consider the following argument.

If x is a real number then x >= 1 since ln(1) = 0.

However the derivative of x^2 is 2x and this is less than the derivative of ln(x) which is 1/x.

2x > 1/x for all x >= 1 and since 1^2 > ln(1) = 0, you will never have a point of intersection for the relationship x^2 = ln(x).

So as long as x is a real number, no solution exists.
 
  • #3
When you have a simple equation like that, you can also plot both sides and realize that the two curves can never cross.
 
  • #4
I should've mentioned this, I'm sorry, I'm looking for a complex solution to this problem. I was already aware that there wasn't a real solution for x.
 
  • #5
karpmage said:
I should've mentioned this, I'm sorry, I'm looking for a complex solution to this problem. I was already aware that there wasn't a real solution for x.

In which case you should define what ln x means, since the complex logarithm is multi-valued. Let's say you mean the principal value.

Let x = a + bi. Then perform the necessary operations on each side of the equation (easier when x is expressed in polar (exponential) form for the RHS), equate real and imaginary parts to get simultaneous equations. Plug that into a numerical solver and see if you get any real solutions for a and b.
 
  • #8
Curious3141 said:
...or you can do that, and just plug the original equation into a numerical solver. :biggrin:

Yeah, I'm getting lazy :blushing:
 
  • #9
DrClaude said:
Yeah, I'm getting lazy :blushing:

Mathspeak for "lazy" is "elegant". :rofl:
 

What is the solution of x^2=ln(x)?

The solution of x^2=ln(x) cannot be expressed in terms of elementary functions and requires the use of a special function known as the Lambert W function.

How do you solve x^2=ln(x)?

To solve x^2=ln(x), you can use the Lambert W function or numerically approximate the solution using a computer program or calculator.

Is there more than one solution to x^2=ln(x)?

Yes, there are two solutions to x^2=ln(x) which can be found using the Lambert W function. These solutions are x=1 and x=e^2, where e is the Euler's number.

What is the relationship between x^2 and ln(x)?

The relationship between x^2 and ln(x) is that they are inverse functions of each other. This means that if you plug in x^2 as the input for ln(x), you will get back the original value of x.

Can x^2=ln(x) be solved algebraically?

No, x^2=ln(x) cannot be solved algebraically using basic operations and functions. The use of the Lambert W function is necessary to find the exact solutions to this equation.

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