A Real Exponential Equation | x^{x+y}=(x+y)^y
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- เผยแพร่เมื่อ 16 ม.ค. 2025
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doesn’t this assume a relationship between y and x? What if they are mutually independent? Or if y is another function of x.
x and y are two arbitrary real numbers with x!=0. Then there must be some value of the quotient y/x, call it k. Then we can rewrite that relation as y=kx. And that new relation is a very useful substitution for this problem.
The real numbers form a field, wich means, among other things, that a multiplication binary operation * is defined and the result is another element inside the field, that exists a unique multiplicative identity called 1, and that any element apart form 0, say x, has an inverse x^-1 that is also a real number such that x * x^-1 = 1.
Let x and y be real numbers. Then x has inverse x^-1 that is also a real number, then if we multiply y * x^-1 the result is another real number, say k: y * x^-1 = k, then, y * x^-1 * x = k * x -> y * 1 = y = k * x.
Conclusion: if x and y are real numbers, then exists a real number k such that y = k * x.
All non zero real numbers have a relationship, they're multiples of one another.
Nice! This y=kx makes this problem so easy!😀💯💯
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
I loved this video!!!❤❤❤
Glad to hear that! 💕
This is one of the most difficult equation I've seen on this channel ...
Two solutions are obvious to me : (x,y) = (2,2) and (1,0)
Then II made a 3d graph of f(x,y) around these, and it seems there are an infinity of solutions going through a curve (the one you present at the end of your video, but failed to find a direct equation describing the curve as y=g(x) or the opposite ...
You did great with parametric couple of equations describing the curve.
But it seems the curve is still incomplete as ... (x,y) = (1,-2) seems to be a solution too ... and it's not on the curve
But all solutions with x+y
You can get (1,-2) with k = -2
@@MizardXYT Yes I know, that's how I found it, but it's not on the curve presented at the end of the video ... and you cannot find the solutions with y=-4,-6,-8,-10... using that method.
In fact, there is no continuous solutions for k < -1, only particular solutions where y = -2n, n natural. And I didn't even find any solution with k integer < -2 (so with x integer too). I don't know if such exist, but I don't think so ..
(1,-2) , (2.603,-4) , (4.561,-6) , (6.612,-8) , (8.866,-10) , (10.710,-12) , (12.743,-14) ...I don't even see a logic with x+y ...
@@tontonbeber4555 For k below -1, only integer values produces real solutions. But for values of k between -1 and 0 you can get values on the lower curve. But I don't think there is any exact formula.
k = -0.590394 => (1.693781, -1)
k = -0.739533 => (2.704405, -2)
k = -0.804935 => (3.727007, -3)
k = -0.842554 => (4.747466, -4)
k = -0.867315 => (5.764912, -5)
k = -0.884984 => (6.779780, -6)
k = -0.898291 => (7.792573, -7)
k = -0.908708 => (8.803703, -8)
k = -0.917105 => (9.813488, -9)
The negative integer solutions does not lie on the curve for -1 < k < 0.
@@MizardXYT Yes, sure I know there are continuous solutions on the curve for any k > -1 including -1 < k < 0.
All the solution you list are on the curve (look at the video, there is a curve for y
@@MizardXYT And of course I wont enter into the discussion about (0,0) ... everybody knows how polemic it could be ^^
Nice zvideo and solution! You may also consider solutions with: 1) k as negative integer smaller than -1 2) k as real with -1< k< 0.
And who said that y is a function of x?
I got (1,0) just by plugging it in.
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
Nice
Thank you. This is a definitely 'see again' video.
Wow, thank you!
why you replace y=kx ?
Isn't (-2,-2) also a real solution that should be obtained with K=1 ?
k=-1 returns no solutions. Other than that, amazing video.
Wow!
k=-1 => y=-x is an asymptote to the curve. There are no solution on the curve with k less or equal than -1
But you can find special additional solutions with k< -1 (which means x+y
1:16: ...=(x+kx)^x. After it, I think the results are wrong.
x=1, y=0
That’s it?
Aren't there complex solutions? What if k= a+ib?
Are there?
График чего от чего строится в конце? И что за экстремальная пара (2.2, 9.18)?
Where are you from? You seem to speak french, but I'm curious about your nationality
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
how do we know we are not missing some solutions? like if y/x=k but two different solutions (x1,y1) and (x2,y2) give the same k(y1/x1=k and y2/x2=k), but plugging in that k into this method can miss some pairs where yi/xi=k? can someone prove the method in the video gives a complete set of solutions.
If *x, y > 0* is a solution, then there is exactly one *k > 0* such that *y = kx* and
*x ^ { (1 + k) * x } = [ (1 + k) * x ] ^ { k * x }*
Using the exact same steps as in this video, it follows *x = (1 + k)^k.* If we restrict ourselves to positive solutions beforehand, none of the steps lead to multiple solutions for *x,* as far as I can tell. Thus each *k* yields exactly one solution *(x; y).*
I find a partial solution when x=y, getting x^2x=(2x)^x so x^x = 2^x or x=2. Thanks for your complete solution 👍
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
1) You are letting *y = x.* You are replacing y with x in your work.
2) It is x^(2x) = (2x)^x. Grouping symbols are needed in *both* places because of the Order of Operations.
One possibility: x=1; y=0,1
On looking at the equation my first thought is to go outside because that's hideous.
When you raise to the 1/x you get +- if x Is even so more cases Need to be examined
Super. Now find all the solutions where
y≠kx.
exactly
Y can be always expressed as kx
Solutions loose continuity or are sparse for y>115 on desmos Weird Why. negative values ok though. y
Must be computational artifacts. Solutions there but harder to calculate.
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
x = 1, y = 0 or x = 2, y = 2
یک معادله با دو مجهول در واقع تابعی از R یعنی پارامتری در R^2 یعنی صفحه است
👍
در یک معادله با دو مجهول همواره باید در فکر یافتن جوابهای پارامتری باشیم
همواره می توان به یک مجهول مقدار داد و در یافتن جواب برای مجهول دیگر باشیم این همه جوابها نیست بلکه تکنیکی برای یافتن جوابهای پارامتری است
Good thinking
احسنت
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
x=y=2
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
@@aymanhindi9610 mind on more with integrals or differential equations or even Laplace or Fourier transforms?