Solving x²=-1/2ln(x) with x in (0,2]

In summary, the equation x²=-1/2ln(x) has complex solutions and cannot be solved using real numbers. To solve it, the complex logarithm function can be used. It is possible to have multiple solutions for x, each in the form of a complex number. However, x must be a real number between 0 and 2. The complex solutions have significance in representing points of intersection between two graphs and can have practical applications in various fields.
  • #1
Paragon
10
0
Hi,:tongue2:

so, I can't solve the embarissing:

[tex] x^{2} = -\frac{1}{2}ln(x) [/tex]

, where [tex] x \in ]0, 2] [/tex] or [tex] (0 < x \geq 2 )[/tex]

any hep would be nice...
thanx for your pacience!
 
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  • #2
Paragon: Equations of this kind can't be solved exactly, i.e the solutions can't be expressed in elementary functions. You can use a numerical method, such as Newton's Method, to approximate the solution.
 
  • #4
Hi, why don't you try with Wright omega function it suits your requirement

y+lny=z
or as suggested by others use LambertW function
 

1. What is the solution to x²=-1/2ln(x) when x is between 0 and 2?

The solution to this equation is complex numbers. It cannot be solved using real numbers.

2. How do you solve x²=-1/2ln(x) using complex numbers?

To solve this equation, you can use the complex logarithm function ln(z) = ln|z| + i arg(z), where i is the imaginary unit and arg(z) is the principal value of the argument of z. You can then substitute the complex logarithm into the equation and solve for x.

3. Is it possible to have multiple solutions to x²=-1/2ln(x)?

Yes, it is possible to have multiple solutions for x in this equation. Each solution will be in the form of a complex number.

4. Can x be any complex number between 0 and 2 in x²=-1/2ln(x)?

No, x cannot be any complex number between 0 and 2. The equation is only defined for x values between 0 and 2, meaning that x must be a real number between 0 and 2.

5. What is the significance of having complex solutions in x²=-1/2ln(x)?

The complex solutions in this equation represent the points where the graph of y=x² and the graph of y=-1/2ln(x) intersect. These solutions can also have real-world applications in fields such as physics and engineering.

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