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

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.

Last edited:

Answers and Replies

chiro
Science Advisor
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.

DrClaude
Mentor
When you have a simple equation like that, you can also plot both sides and realize that the two curves can never cross.

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.

Curious3141
Homework Helper
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.

DrClaude
Mentor
...or you can do that, and just plug the original equation into a numerical solver.
Yeah, I'm getting lazy

Curious3141
Homework Helper
Yeah, I'm getting lazy
Mathspeak for "lazy" is "elegant". :rofl: