Solving x^y=y^x & x+y=6 with Deduction

  • Thread starter arka.sharma
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In summary, the conversation discusses a set of equations and the possibility of finding a deductive way to solve them. The equations are x^y = y^x and x+y = 6, and the possible solutions are (x=4,y=2),(x=3,y=3), and (x=2,y=4). The conversation also mentions the equation (6-y)^y == y^(6-y) and suggests solving it graphically or using a numerical iteration method. It is concluded that there is no elementary algebraic method to solve this equation.
  • #1
arka.sharma
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0
Hi All,

I was given two equations back in my school days to solve for both x and y it is as follows

x^y = y^x & x+y = 6

Now it can be seen that following would be possible solutions (x=4,y=2),(x=3,y=3),(x=2,y=4).
But is there any deductive way to solve this ?

Regards,
Arka
 
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  • #2
Solving (6-y)^y == y^(6-y)
Which can be represented graphically.
 
  • #3
Thanks for your reply. But how to solve the equation (6-y)^y == y^(6-y) ?
 
  • #4
Put the RHS and the LHS of the equation on a graphic and see where these curve cut each other.
You will find the root that you already know.
If there are other roots, you will likely find out.
Afterward, you might find out arguments to "prove" you found all the roots.
The graphical representation will help a lot.
 
  • #5
arka.sharma said:
Thanks for your reply. But how to solve the equation (6-y)^y == y^(6-y) ?
You were told to solve it graphically. It can also be solved using a numerical iteration method. If you are looking for an elementary algebraic way of solving it, there is none.
 

1. What is the equation "x^y=y^x" used for?

The equation "x^y=y^x" is used for determining the relationship between two variables, x and y, when they are raised to different powers.

2. How do you solve for x and y in the equation "x^y=y^x"?

To solve for x and y in the equation "x^y=y^x", we can use the method of logarithms. Taking the logarithm of both sides, we get y*log(x) = x*log(y). By rearranging the terms, we can solve for x in terms of y, and then substitute this value into the equation x+y=6 to solve for y. Similarly, we can solve for y in terms of x, and substitute this value into the equation x+y=6 to solve for x.

3. What is the significance of solving "x^y=y^x" and "x+y=6" together?

Solving "x^y=y^x" and "x+y=6" together allows us to find the specific values of x and y that satisfy both equations simultaneously. This can be useful in various applications, such as in determining the optimal ratio of two quantities or in solving problems involving exponential and linear relationships.

4. Can the equations "x^y=y^x" and "x+y=6" be solved without using deduction?

Yes, these equations can be solved without using deduction. Other methods, such as substitution or graphing, can also be used to find the solutions. However, using deduction can be a more efficient and systematic approach to solving these equations.

5. Are there any limitations or conditions for solving "x^y=y^x" and "x+y=6" with deduction?

Yes, there are some limitations and conditions for solving "x^y=y^x" and "x+y=6" with deduction. For example, both equations must be true for the same values of x and y, and the solutions must satisfy the domain and range restrictions of exponential and logarithmic functions. Additionally, the solutions may not always be unique, as there may be multiple values of x and y that satisfy both equations.

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