Finding Solutions to y^x = x^y

  • Thread starter jcsd
  • Start date
In summary, the equation y^x = x^y has several trivial solutions, such as y = x and a set of non-trivial solutions for y^{y^y} = y^{y^2}. There exists a method that can be used to obtain a general solution, which has been seen and used by some users at scienceforums.net. Although the method and its name are not remembered, it has led to a family of non-trivial solutions in the real numbers. In the complex domain, things become more complicated due to the existence of multiple branches in the natural logarithm, but there are likely additional complex solutions.
  • #1
jcsd
Science Advisor
Gold Member
2,115
13
For the equation:

[tex]y^x = x^y[/tex]

There are several trivial solutions:

such as:

[tex]y = x[/tex]

and a further set of trivial solutions for:

[tex]y^{y^y} = y^{y^2}[/tex]

Now I know for a fact there are also sevral non-trivial solutions and there is a method thta could possibly obtain a
general solution, so does anyone here know how to obtian a general solution or if not suggest other sets of solutions?
 
Mathematics news on Phys.org
  • #2
can you show the method that possibly give a general solution?

i know for sure that some users at scienceforums.net will be glad to know the solution.
 
  • #3
I have seen the method before and a few non-trivial solutions obtained from it, but unfortunately I cannot remember it or even it's name.

As I said before their most definitely does exist non-trivial solutions to this equation which some might be interetsed to know.
 
  • #4
Originally posted by jcsd
For the equation:

[tex]y^x = x^y[/tex]

does anyone here know how to obtian a general solution or if not suggest other sets of solutions?

[tex]y^x=x^y[/tex]
[tex]xlny=ylnx[/tex] (assuming x and y are > 0 )
[tex]\frac{ln x}{x}=\frac{ln y}{y}[/tex] (x and y are non-zero by assumption above)
[tex]x^{\frac{1}{x}}=y^{\frac{1}{y}}[/tex]

Now , in the reals, for positive [tex]x[/tex]
[tex]f(x)=x^{\frac{1}{x}}[/tex] has two inverses for any [tex]f(x) \in (1,e^{\frac{1}{e}})[/tex] which leads to a family of non-trivial solutions. For example:
2 and 4 or 3 and 2.47779..

Fractional roots in the complex domain are multi-valued, so things get a bit more complicated if this approach is applied for x,y < 0.
[tex]x^y=y^x[/tex]
[tex]e^{ylnx}=e^{xlny}[/tex]
[tex]ylnx=xlny + n2\pi i[/tex]
Since the natural long in complex numbers has multiple branches, things get a bit sticky here, but there are probably additional complex solutions.
 

1. What is the equation "y^x = x^y" trying to solve for?

The equation "y^x = x^y" is trying to solve for the values of x and y that satisfy the equation. In other words, it is looking for the specific values of x and y that make the left and right sides of the equation equal.

2. Is there a straightforward solution to the equation "y^x = x^y"?

No, there is no straightforward solution to this equation. It is a complex equation and does not have a simple algebraic solution. Instead, it requires the use of advanced mathematical techniques, such as logarithms, to find a numerical solution.

3. Can you explain why finding solutions to "y^x = x^y" is important?

The equation "y^x = x^y" has many real-world applications, such as in thermodynamics, economics, and biology. Solving this equation can help us understand the relationships between different variables and make predictions about their behavior. Additionally, finding solutions to this equation can lead to new insights and discoveries in various fields of science and mathematics.

4. Are there any known solutions to "y^x = x^y"?

Yes, there are some known solutions to this equation. For example, when x = y, the equation becomes "y^y = y^y", which is always true for any value of y. Another known solution is when x = 2 and y = 4 (or vice versa), which satisfies the equation "4^2 = 2^4". However, these are only a few of the infinite solutions to this equation.

5. Can this equation be solved for all values of x and y?

No, this equation cannot be solved for all values of x and y. Some values may not have a real solution, while others may have an infinite number of solutions. Additionally, finding solutions for large or irrational values of x and y can be extremely difficult, if not impossible, to find. Therefore, it is important to understand the limitations and complexities of this equation when attempting to find solutions.

Similar threads

Replies
1
Views
700
  • General Math
Replies
2
Views
687
Replies
2
Views
1K
Replies
4
Views
736
  • General Math
Replies
13
Views
1K
Replies
3
Views
1K
Replies
11
Views
766
  • Calculus and Beyond Homework Help
Replies
7
Views
642
  • General Math
Replies
16
Views
3K
Replies
13
Views
1K
Back
Top