I'm little bit stuck with this system of equations : x^y=y^x and x^3=y^2 An obvious solution is (x,y)=(1,1) but what about the solution (9/4,27/8) ? I know the relation a^r=e^(r log a) but it doesn't help me.Thanks

Lorelei Patterson

Lorelei Patterson

Answered question

2022-07-21

How to solve this kind of equation ( x y = y x ) ( x y = y x )
I'm little bit stuck with this system of equations :
x y = y x and x 3 = y 2
An obvious solution is ( x , y ) = ( 1 , 1 ) but what about the solution ( 9 / 4 , 27 / 8 )?
I know the relation a r = e r log a but it doesn't help me.
Thanks

Answer & Explanation

thenurssoullu

thenurssoullu

Beginner2022-07-22Added 13 answers

Besides the easy solution x = y = 1 you can do the following:
Take log on both sides of both equations to have:
3 log x = 2 log y , y log x = x log y .
Divide side by side (and here you should note that x = y = 1 is a solution and hence, we cannot divide by log x or log y , otherwise we have the following), to come up with:
x y = 2 3 .
Divide side by side (and here you should note that x = y = 1 is a solution and hence, we cannot divide by log x or log y , otherwise we have the following), to come up with:
x y = 2 3 .
Since the equation x 3 = y 2 can also be written as: x 2 x / y 2 = 1 , provided y 0, we have that:
( x y ) 2 x = 1 x = 9 4 ,
which leads to the desired result.
Hope this helps.
Cheers!
Almintas2l

Almintas2l

Beginner2022-07-23Added 6 answers

Note that the second equation gives x 3 k = y 2 k for arbitrary k and divide the first equation by this to get
x y 3 k = y x 2 k
Now choose k = y 3 so that 1 = y x 2 k and either y = 1 or x = 2 k
This gives x = y = 1 or ( 2 k ) 3 = ( 3 k ) 2 which gives k = 0 (anomalous solution x = y = 0) or k = 9 8

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