Sir, before you start solving, you have to write x ≠ 0, y ≠ 0. After, when you write y = kx, you have to write k ≠ 0. So, if we consider x=1 and y=1 , that is a solution. And if y=kx, k=1 in this case. Your solution leads to x = k^(1/(k-1) and y = k^(k/(k-1). This would seem to mean that k cannot be 1. And yet x=y=1 is indeed a solution. Finally, as x and y are only expressed in terms of k, a constant, then there is indeed an infinity of possible answers, and not only 3. The structure of the development of the problem is good, but it is therefore missing very decisive elements of answers.
@@ggaskoin this is not what I leant, I'm currently student in a french maths school. They taught us that by convention, 0^0 is 1. It would be awkward to me if this convention applied only in France... Moreover, if you see it as the limit of x^x in 0+, 1 is the only answer, if you want it to respect the rule x^0 = 1 whatever x is, 1 is also the only option.
@@leto7490 you admit that it is a "convention". And I just told you that it depends on the context. Sometimes it is treated as "1" and sometimes as undefined as intuitively 0 power by anything can't give you any number other than 0. Nevertheless the solution brought by video is not correct because you can't just assume that y=kx out of the blue as how do you know that dependency is linear? You don't. It is more complex to solve this and as math student you probably know that :)
@@ggaskoin @Michał G Well mister challenger. Take two real numbers x and y, suppose x is not 0. There is a real number k such that y = kx. Every maths student knows that. k is even equal y/x, which is defined because x ≠ 0. Incredible. Then, yes it is a convention, among the mathematicians, so it is true. I don't knox if you know that maths are built on unprovable properties.
There is an infinite number of solutions, when x == y which is kind of obvious, possibly x = y = 0 may not be allowed, with this one exception, any x == y will work.
I solved it easily in my mind: - As long as "x=y", "{x'y} = R`{0}" - Neither "y" nor "x" are equal to "0" - "x={4||2}" and "y={2||4}" are the only N. numbers combination where "x≠y". This is at least what can derived by logic (and no calculation is needed for). 👍✨️
This is not a general solution for the problem! From the beginning you can derive lnx/x= lny/y (for x and y > 0 obviously). Now the question is, is the function f(x) = lnx/x a bijection, i.e. for 1 value of f(x), is there one and only value of x. After analyzing the function you see that it is not: its limit for x--> 0 is "- infinite", it has a 0 value for x=1, a local maximum f(x max)= 1/e for x max=e, and it decreases from x= e to x = "+ infinite" where its limit is 0. Therefore f is a bijection for x between 1 and e, and separately it is one for x between e and "+ infinite". Therefore, for any constant value Z of f(x) comprised between 0 and 1/e, there will be 2 solutions to the equation lnx / x = k --> the original equation has an infinity of solutions, and for each x comprised between 1 and e= 2.7... there is an y comprised between e= 2.7 and + infinite. Nevertheless I could not find yet an analytical solution, and this is not satisfying.
But one can check in excel that for all values of K from 1.01 to 1000, the parametric functions y= k^(1/(k-1)) and x= k^(k/(k-1)) verify the initial condition x^y=y^x! But numerical solutions are not very satisfying...
problem is, for solutions aside trivial y=x, you need lambert function. If going the parametric way like in the vid, reaching finally a y=f(x) expression at some point requires inverse function of x=k^(1/(k-1)) -> known problem: you cannot find inverse function to x*e^x in algebraic way. Typically lambert function is introduced in this case.
I only missed comments from topologists here, math is fun and the critics are healthy in my opinion. I would avoid the square root in this situation though as one can obtain X=k^(1/k-1) without it.
the obvious answer is when x=y but if x doesn't equal y then the obvious answers are 2 and 4 because 2^4 is indeed equal to 4^2. I don't know about the solution that you found.
It seems like you didnt even watched the whole video. His answer was an infinite set of numbers for this problem. Which is much better answer than yours. But probably you didnt want to learn anything new. I mean, it is good that you can see a couple of answers right away but there is always room for improvement.
You're partially right, he actually did quite a few things carelessly, like assuming that X and Y where strictly positive (the case X=0 or Y=0 isn't difficult to treat at all, it is easy to show that in this case the other unknown is equal to 0 as well but this isn't said in the video) and then dividing by K-1 which excluded the case K=1 and X=Y. The rest is correct even if there are a few useless steps, and and if you watch carefully and plug in his formula K=2 you'll get X=2 and Y=4. You can also make sens out of exponentiation when the number you exponentiate is negative, as long as the exponent is an integer. In this case the solutions X=Y when X and Y are negative integers work, but these aren't the sole solutions since X=-2, Y=-4 works as well. In facts, if X, Y are 2 positive integers that solve this equation and have the same parity, then -X, -Y also solve this equation. Actually, I am not able to describe precisely the set of solutions of the whole real line. I can't give you an explicit solution but it is even possible to show with the intermediate value theorem that for any strictly positive pair number Y there exist a real strictly negative number X so that X, Y solve the equation, and that's pretty much all the solutions of the real line.
@@robertogutierrez3290 yes, he found an infinite set of solutions; but those are not all solutions! And he did not mention that this solution is undefined when k=1. And if you swap the expressions fof x and y, this would also be another infinite set of solutions. Besides, there is at least one more infinite set: x=y > 0. Besides, there was absolutely no need to extract the square root ...
After watching this video I can up with a few answers: 1. When X=Y, but: When X=Y=0 it's more complicated. When you put it in an equation 0^0 = 0^0 then it's correct as on both sides you have the same thing. But when you take into the consideration the value of 0^0 which is either 0 (0 to any power is 0) or 1 (any number to power 0 is 1) then it fails as you can get: 0=0, 0=1 or 1=1 (That's why 0^0 is undefined). 2. When X≠Y The final answers that he achieved in this video are correct I think (I haven't done much of calculations to Chech it) but he should add one more thing that K≠1 as otherwise he would have to divide by 0 which is undefined as well (both infinity and negative infinity). To whoever reads it - yes math is difficult but it's a philosophy. Everything gives you an answer depending on how you encounter the problem. Hope you have fun reading this :)
The complete set of solutions would be like this: (x,y)=(k^(1/1-k); k^(k/k-1) for every k>0 and different than 1. And also (x,y)=(a,a) for every integer “a”different than 0
One could summarize the result as follows: in the {x , y }-plane there are two solution curves :(1) the curve in parametric form as {x =k^(1/(k-1)),y = k^(k/(k-1)) } , and the line y=x , x>0.
1) X > 0 Y > 0. 2) Take the xy-th root of both sides. X^(1/x) = Y^(1/y) ergo X = Y. 3) There are other solutions when X and Y have certain relationships, e.g. 2 and 4.
0:40 What's the point of taking the square root here? It would seem to me that we can just skip the square root entirely but otherwise follow the same logic as you do. Basically, we have x^(kx)=(kx)^x which we can also express as (x^k)^x=(kx)^x. Taking both sides to the power of 1/x we get x^k=kx which is the same as you get at 3:00. Taking the square root seems completely superfluous.
interesting technique (y =kx); however, some restrictions should be considered along the process: x cannot be zero because at some point you divide exponents by x; k cannot be 1 (otherwise the denominator k-1 become zero), and so on. I liked this video.
@Carlos Carlos because they're invalid or excessively complicated. Y is a multiple of X in all solutions, therefore Y=kX For example if Y= X sin X, then sin X can just be k, therefore Y=kX
@@Jon-ov4nc Wrong. sin(x) could not be k because k is supposed to be a constant. Also, for k=1, y=kx become y=x which obviously also solves the problem, albeit in a trivial manner.
Looks like you asked to figure y = f(x) dependence, but provided just one pair of values instead - when even for x = sqrt (3) there are two possible y-s - sqrt(3) and 3*sqrt(3)
Exponentiation is not well-defined for negative, fractional bases. If a is not an integer, the value of a^b is defined to mean exp(b ln a). And ln a is not defined for negative values of a.
If x > 1 and y > 1, then besides x = y ≠ e, there is another x, y pair where x ≠ y that satisfies the conditions of the equation. In the event that 1 < x < e, there is a y > e such that x^(1/x) = y^(1/y) which is equivalent to x^y = y^x. See what I mean? Note that if x = y = e, then there is not another pair x, y that satisfies the conditions of the equation if it must be the case that x > 1 and y > 1.
If k=9 then: x=k^(1/(k-1)=9^[1/(9-1)]= 9^(1/8), and the same with y=k^[9/(9-1)], next step to take k=9+1=10 repead as before . Baase on mathematical analysis theorem we can state: there are an infinite numbers of pairs of numbers satisfying these conditions.
x = y = 1 ; but generally x = y for All Real values x & y, except 0; e.g., x = y = 3 Check: 3^3 = 3^3 = 27 = 27 e.g., x = y = -1/2 Check: (-1/2)^(-1/2) = (-1/2)^(-1/2) = -i sqrt(2) Any Real Number works, except 0, unless you accept 0^0 = 1, formulaically it works, but what does 0^0 mean? It may be undefined, one of Gödel’s inconsistencies?
@@rogeriodepaulavalvano1541 it is not, 0^0 is 1 there are interesting videos proving it. He also ignores the possibility of x and y being equal to 1 or any other x=y possibility
@@5reck0 0^0 is not 1. There are no videos "proving" that. Of necessity, it is an indeterminate form. Consider the behavior of 0^y vs that of x^0 along the y and x axes respectively. One is constant at 1 and the other is constant at 0. One can adopt a convention of setting 0^0 = 1, which is convenient in some fields of mathematics (combinatorics, algebra) but that's a different thing than "proving" that the formula is true in any natural way.
@@rickdesper bro 0^0 = 1 We know that (anything) counted zero times is 1 And zero counted any number is 0 0 counted 0 times is not 0 times anything because there is no 0 So its 1 If dont like logical proofs go instead search up a 0^0 mathematical proof its like 7 ligns its not really complicated
As a try to solve the equation in integer numbers, this could be a good one, once you set the appropriate restrictions, like the ones commented below. But using an implicit plotter, one can observe that not only y=x is part of the (implicit) curve, but also a double pair of hyperbolic-type curves; 2 in the I+III quadrants, and another pair in the II+IV quadrants. So, the union of the five curves (the 4 symmetric of hyperbolic type, plus the line y=x) shows that there exist more solutions than the integers (or even real numbers) populating the y=x. I wonder if there exists an algebraic method to find the solutions that lies in those 4 hyperbolic curves... 😮🤨
tried my very best. obviously y=x is trivial solution to the problem call it algebraic or anyhow. But about the further solutions: Finally the video gives (I) x=k^(1/(k-1) and (II) y=k^(k/(k-1) still containing the auxiliary constant k. if you could transform equation (I) to k=f(x) and insert this term into equation (II) or vice versa, you had y=g(x) as further solution series aside trivial solution y=x. Though I failed to find inverse function of (I). inverse function of x^x (or x*e^x) seems impossible to be found algebraically. Does someone know better?
@@peterhannershuber6420 One could try to generalize by taking y = f(x) where f(x) is _any_ chosen function. For _some_ functions f the equation x^f(x) = f(x)^x (😱 ) would in principle have numerical solutions. But then for every such solution we could calculate the number k = y/x.
if you want to know about more property of x and y , you can draw the graph f(x)= ln(x) / x. Right side is result that you can get from taking log to both sides of question and arrange for same letter. Such as ln(y)/y=ln(x)/x.
Great explanation, but roots x € R and y € R when x equals y are missed. Good video. Didn't even think that there could be way of square roots and much more exponents.
In the problem the condition that x must be different from y is not set, so, it follows that in the domain of natural numbers there are countless solutions - (x=1, y=1); (x=2, y=2), (x=3, y=3)...
Taking square roots was totally pointless! 😉 y = kx gives x^(kx) = (kx)^x That is, (x^k)^x = (kx)^x. → x^k = kx → x^(k-1) = k → x = k^(1/(k-1)), y = k^(k/(k-1)).
couldn't you just use the x-th root (to the power of 1/x) to get rid of the x-exponents at 0:56 when you did it at 2:43 but with the extra unnecessary steps in-between?
first: to my limited math knowledge (I'm just a chemist not a mathematician), one equation containing two variables always has infinite solutions. Second: I clearly agree with all the comments below about y=x=0 beeing a solution, as for any y=x would be trivial solution to the problem, as well as obviously the problem of k=1 is totally ignored in the video. Third: my question: after you solved x=k^(1/(k-1) and y=k^(k/(k-1), why don't you proceed to find a direct term of y=f(x) to at least determine the solution series given for the case you followed previously?
To be clear, what you're showing is that, for any value of k=/=1, k > 0, there is a pair (x,kx) that satisfy the equation x^kx = (kx)^x. (For k=1, this is true for any positive value of x.) You're not solving an equation so much as you are defining a function.
If k=2 then x=2 and y=4 and its true. If k=1 then x=1^(1/0) and y=1^(1/0) and both of them are indeterminated and equal, so x=y. The solution in video is universal
How can you which is a dependant variable and which is an independent variable. By assuming y=kx u are assuming them to related without any basis. 1 eq’n 2variables so this questions incomplete
I have a very very simple solution to this problem - and can be down without pen and paper. Simple huristics. One ( if not the only ) solution is that x and y are interchangable. so - x=y. anyone want to prove me wrong? Let me add this. You have to have a match between the number of variable and equations to get a real answer. Since we have one equation, there can be no definative answer to this problem. You can get a releationship between x and y, but no definiative answer ( no x = value and no y = value ).
I need to ask, right where you do the square root, how are you sure that x^kx is positive? Because if it isn't then you can't really do the square root, this means the solutions you will fet from there may or may not be the same as the initial
Lovely algebra, but you can't add a specific equation to the original problem. There is no single answer because there is only one equation for two variables. Thanks for the attempt, anyway!
This is not right, why k would be a constant then y and x will have a linear relationship which is assumed and again later we see k is a function of x that means it could not have been a constant, a self contradicting exercise
I don't think, step at 2:51 is valid. Multiplying the powers with its reciprocal? Which rule is that? I think it's invalid process to solve. Correct me please.
My friend you presented one equation with 2 unknown which is unsolvable, you inserted a second equation out of thin air, which is absolutely no part of the solución, you said y=kx, this is a second equation. I will ask you to solve it using y=sin(px). If you present a problem of this type you need to give both equations at the beginning no after the facts
The assumption of y=kx is wrong as this will restrict the solution of the original equation. In fact, the original equation has infinite solutions because one equation with two unknowns does not have unique solution!!
@@Axenvyy This is basic vector algebra, one equation two unknowns has infinite solutions!! For example, x = 1 = y, x = 2 = y, etc., all satisfy the equation!
@@rclrd1This is already an assumption that only two numbers satisfying this condition are true in the subsequent derivation. That means an extra condition is added to qualify the x and y relation!!
Sir, before you start solving, you have to write x ≠ 0, y ≠ 0. After, when you write y = kx, you have to write k ≠ 0. So, if we consider x=1 and y=1 , that is a solution. And if y=kx, k=1 in this case. Your solution leads to x = k^(1/(k-1) and y = k^(k/(k-1). This would seem to mean that k cannot be 1. And yet x=y=1 is indeed a solution. Finally, as x and y are only expressed in terms of k, a constant, then there is indeed an infinity of possible answers, and not only 3. The structure of the development of the problem is good, but it is therefore missing very decisive elements of answers.
I agree with everything but the beginning. x=y=0 is a solution since 0^0 = 1
@@leto7490depends on the context. In analysis it is undefined..
@@ggaskoin this is not what I leant, I'm currently student in a french maths school. They taught us that by convention, 0^0 is 1. It would be awkward to me if this convention applied only in France...
Moreover, if you see it as the limit of x^x in 0+, 1 is the only answer, if you want it to respect the rule x^0 = 1 whatever x is, 1 is also the only option.
@@leto7490 you admit that it is a "convention". And I just told you that it depends on the context. Sometimes it is treated as "1" and sometimes as undefined as intuitively 0 power by anything can't give you any number other than 0. Nevertheless the solution brought by video is not correct because you can't just assume that y=kx out of the blue as how do you know that dependency is linear? You don't. It is more complex to solve this and as math student you probably know that :)
@@ggaskoin @Michał G Well mister challenger. Take two real numbers x and y, suppose x is not 0. There is a real number k such that y = kx. Every maths student knows that. k is even equal y/x, which is defined because x ≠ 0. Incredible.
Then, yes it is a convention, among the mathematicians, so it is true. I don't knox if you know that maths are built on unprovable properties.
If x=k, y=k...
x^y=y^x
k^k= k^k
So if x=y then we have a right solution
Also if x=1 and y=√1 we have a solution
x^y=y^x => 1^√1=√1^1 => 1 = 1
And generally, if x=a and y=√a then a√a = √a^a => a^a^1/2 = a^1/2^a => a^a/2 = a^a/2
There is an infinite number of solutions, when x == y which is kind of obvious, possibly x = y = 0 may not be allowed, with this one exception, any x == y will work.
X == Y? This is maths, not programming.
I solved it easily in my mind:
- As long as "x=y", "{x'y} = R`{0}"
- Neither "y" nor "x" are equal to "0"
- "x={4||2}" and "y={2||4}" are the only N. numbers combination where "x≠y".
This is at least what can derived by logic (and no calculation is needed for). 👍✨️
This is not a general solution for the problem!
From the beginning you can derive lnx/x= lny/y (for x and y > 0 obviously).
Now the question is, is the function f(x) = lnx/x a bijection, i.e. for 1 value of f(x), is there one and only value of x. After analyzing the function you see that it is not: its limit for x--> 0 is "- infinite", it has a 0 value for x=1, a local maximum f(x max)= 1/e for x max=e, and it decreases from x= e to x = "+ infinite" where its limit is 0. Therefore f is a bijection for x between 1 and e, and separately it is one for x between e and "+ infinite". Therefore, for any constant value Z of f(x) comprised between 0 and 1/e, there will be 2 solutions to the equation lnx / x = k --> the original equation has an infinity of solutions, and for each x comprised between 1 and e= 2.7... there is an y comprised between e= 2.7 and + infinite.
Nevertheless I could not find yet an analytical solution, and this is not satisfying.
But one can check in excel that for all values of K from 1.01 to 1000, the parametric functions y= k^(1/(k-1)) and x= k^(k/(k-1)) verify the initial condition x^y=y^x! But numerical solutions are not very satisfying...
problem is, for solutions aside trivial y=x, you need lambert function. If going the parametric way like in the vid, reaching finally a y=f(x) expression at some point requires inverse function of x=k^(1/(k-1)) -> known problem: you cannot find inverse function to x*e^x in algebraic way. Typically lambert function is introduced in this case.
As long as x=y you can have infinite number of roots.
2^4=4^2
I only missed comments from topologists here, math is fun and the critics are healthy in my opinion. I would avoid the square root in this situation though as one can obtain X=k^(1/k-1) without it.
the obvious answer is when x=y but if x doesn't equal y then the obvious answers are 2 and 4 because 2^4 is indeed equal to 4^2. I don't know about the solution that you found.
My answer if x=2,y=4 , if x=4,y=2
It seems like you didnt even watched the whole video. His answer was an infinite set of numbers for this problem. Which is much better answer than yours. But probably you didnt want to learn anything new. I mean, it is good that you can see a couple of answers right away but there is always room for improvement.
You're partially right, he actually did quite a few things carelessly, like assuming that X and Y where strictly positive (the case X=0 or Y=0 isn't difficult to treat at all, it is easy to show that in this case the other unknown is equal to 0 as well but this isn't said in the video) and then dividing by K-1 which excluded the case K=1 and X=Y. The rest is correct even if there are a few useless steps, and and if you watch carefully and plug in his formula K=2 you'll get X=2 and Y=4. You can also make sens out of exponentiation when the number you exponentiate is negative, as long as the exponent is an integer. In this case the solutions X=Y when X and Y are negative integers work, but these aren't the sole solutions since X=-2, Y=-4 works as well. In facts, if X, Y are 2 positive integers that solve this equation and have the same parity, then -X, -Y also solve this equation. Actually, I am not able to describe precisely the set of solutions of the whole real line. I can't give you an explicit solution but it is even possible to show with the intermediate value theorem that for any strictly positive pair number Y there exist a real strictly negative number X so that X, Y solve the equation, and that's pretty much all the solutions of the real line.
You are zoba
@@robertogutierrez3290 yes, he found an infinite set of solutions; but those are not all solutions!
And he did not mention that this solution is undefined when k=1.
And if you swap the expressions fof x and y, this would also be another infinite set of solutions.
Besides, there is at least one more infinite set: x=y > 0.
Besides, there was absolutely no need to extract the square root ...
After watching this video I can up with a few answers:
1. When X=Y, but:
When X=Y=0 it's more complicated.
When you put it in an equation 0^0 = 0^0 then it's correct as on both sides you have the same thing.
But when you take into the consideration the value of 0^0 which is either 0 (0 to any power is 0) or 1 (any number to power 0 is 1) then it fails as you can get: 0=0, 0=1 or 1=1 (That's why 0^0 is undefined).
2. When X≠Y
The final answers that he achieved in this video are correct I think (I haven't done much of calculations to Chech it) but he should add one more thing that K≠1 as otherwise he would have to divide by 0 which is undefined as well (both infinity and negative infinity).
To whoever reads it - yes math is difficult but it's a philosophy. Everything gives you an answer depending on how you encounter the problem. Hope you have fun reading this :)
The complete set of solutions would be like this: (x,y)=(k^(1/1-k); k^(k/k-1) for every k>0 and different than 1. And also (x,y)=(a,a) for every integer “a”different than 0
One could summarize the result as follows: in the {x , y }-plane there are two solution curves :(1) the curve in parametric form as {x =k^(1/(k-1)),y = k^(k/(k-1)) } , and the line y=x , x>0.
1) X > 0 Y > 0. 2) Take the xy-th root of both sides. X^(1/x) = Y^(1/y) ergo X = Y. 3) There are other solutions when X and Y have certain relationships, e.g. 2 and 4.
0:40 What's the point of taking the square root here? It would seem to me that we can just skip the square root entirely but otherwise follow the same logic as you do. Basically, we have x^(kx)=(kx)^x which we can also express as (x^k)^x=(kx)^x. Taking both sides to the power of 1/x we get x^k=kx which is the same as you get at 3:00. Taking the square root seems completely superfluous.
interesting technique (y =kx); however, some restrictions should be considered along the process: x cannot be zero because at some point you divide exponents by x; k cannot be 1 (otherwise the denominator k-1 become zero), and so on. I liked this video.
K = 1 has valid solution though
X = 1^(1/1-1) = 1^(1/0) = 1^0 = 1
Y = 1^(1/1-1) = 1^0 = 1
X^Y = Y^X, 1^1=1^1
solution of k=0
X = 0^(1/0-1) = 0^(-1/1) = 0
Y = kX = 0
X^Y = Y^X, 0^0=0^0, 1=1
Sort of, it's 'undefined' but still valid
@Carlos Carlos because they're invalid or excessively complicated. Y is a multiple of X in all solutions, therefore Y=kX
For example if Y= X sin X, then sin X can just be k, therefore Y=kX
@@Jon-ov4nc Wrong. sin(x) could not be k because k is supposed to be a constant.
Also, for k=1, y=kx become y=x which obviously also solves the problem, albeit in a trivial manner.
@Fircasice I never said was... Carlos did
Looks like you asked to figure y = f(x) dependence, but provided just one pair of values instead - when even for x = sqrt (3) there are two possible y-s - sqrt(3) and 3*sqrt(3)
Exponentiation is not well-defined for negative, fractional bases. If a is not an integer, the value of a^b is defined to mean exp(b ln a). And ln a is not defined for negative values of a.
If x is not equal to y then
x=2,y=4
x=4،y=2
If x=y
Then many solution is possible
If x > 1 and y > 1, then besides x = y ≠ e, there is another x, y pair where x ≠ y that satisfies the conditions of the equation. In the event that 1 < x < e, there is a y > e such that x^(1/x) = y^(1/y) which is equivalent to x^y = y^x. See what I mean? Note that if x = y = e, then there is not another pair x, y that satisfies the conditions of the equation if it must be the case that x > 1 and y > 1.
Thanks I like your video
For natural numbers only, this problem becomes so much more elegant. The proof that (2,4) and (4,2) are the only solutions is actually very beautiful
The only solution other than x=y.
@@ivanviehoff6025 YES! And in the posted problem, there was no requirement for X ≠ Y nor was there a requirement to find EVERY solution.
If k=9 then: x=k^(1/(k-1)=9^[1/(9-1)]= 9^(1/8), and the same with y=k^[9/(9-1)], next step to take k=9+1=10 repead as before . Baase on mathematical analysis theorem we can state: there are an infinite numbers of pairs of numbers satisfying these conditions.
At a glance, the only way I can see this equation being true is if X=Y
Therefore any value of X will work
But there are an infinite number of less obvious solutions. That's the point!
The solution is actually a linear function of y=x
1,1 2,2 3,3 4,4 etc
2^4=4^2
Why "2^4=4^2"? You have symmetric x^y=y^x
1,1 - 1^1=1^1
2,2 - 2^2=2^2
so on
x = y = 1 ; but generally x = y for All Real values x & y, except 0; e.g., x = y = 3
Check: 3^3 = 3^3 = 27 = 27
e.g., x = y = -1/2
Check: (-1/2)^(-1/2) = (-1/2)^(-1/2) = -i sqrt(2)
Any Real Number works, except 0, unless you accept 0^0 = 1, formulaically it works, but what does 0^0 mean? It may be undefined, one of Gödel’s inconsistencies?
at 2:45 you can do that only for non-zero x. So you only need to treat x=0 as a special case (x=0, y=0 is also a solution)
but 0^0 is an indeterminacy
@@rogeriodepaulavalvano1541 it is not, 0^0 is 1 there are interesting videos proving it. He also ignores the possibility of x and y being equal to 1 or any other x=y possibility
And x=y is also a solution
@@5reck0 0^0 is not 1. There are no videos "proving" that. Of necessity, it is an indeterminate form. Consider the behavior of 0^y vs that of x^0 along the y and x axes respectively. One is constant at 1 and the other is constant at 0. One can adopt a convention of setting 0^0 = 1, which is convenient in some fields of mathematics (combinatorics, algebra) but that's a different thing than "proving" that the formula is true in any natural way.
@@rickdesper bro 0^0 = 1
We know that (anything) counted zero times is 1
And zero counted any number is 0
0 counted 0 times is not 0 times anything because there is no 0
So its 1
If dont like logical proofs go instead search up a 0^0 mathematical proof its like 7 ligns its not really complicated
As a try to solve the equation in integer numbers, this could be a good one, once you set the appropriate restrictions, like the ones commented below. But using an implicit plotter, one can observe that not only y=x is part of the (implicit) curve, but also a double pair of hyperbolic-type curves; 2 in the I+III quadrants, and another pair in the II+IV quadrants. So, the union of the five curves (the 4 symmetric of hyperbolic type, plus the line y=x) shows that there exist more solutions than the integers (or even real numbers) populating the y=x. I wonder if there exists an algebraic method to find the solutions that lies in those 4 hyperbolic curves... 😮🤨
tried my very best. obviously y=x is trivial solution to the problem call it algebraic or anyhow. But about the further solutions: Finally the video gives (I) x=k^(1/(k-1) and (II) y=k^(k/(k-1) still containing the auxiliary constant k. if you could transform equation (I) to k=f(x) and insert this term into equation (II) or vice versa, you had y=g(x) as further solution series aside trivial solution y=x. Though I failed to find inverse function of (I). inverse function of x^x (or x*e^x) seems impossible to be found algebraically. Does someone know better?
@@peterhannershuber6420 "The Union of the Five Curves" sounds like a bad SciFi B-movie
@@peterhannershuber6420 This inverse function of x*e^x is the Lambert W function but it cannot be expressed algebraically.
@@peterhannershuber6420
One could try to generalize by taking y = f(x) where f(x) is _any_ chosen function. For _some_ functions f the equation x^f(x) = f(x)^x (😱 ) would in principle have numerical solutions. But then for every such solution we could calculate the number k = y/x.
x ^ (1/x) = y ^ (1/y)
This is a monotonically decreasing function and x = y is only solution
If you make x and y the same they have infinite correct results. If you make x and y both equal 5 for example, the equation is true as 5^5 = 5^5.
That just means x = y is a solution but it certainly isn't the only one.
Why the square root in the middle of everything? Serves no purpose.
if you want to know about more property of x and y , you can draw the graph f(x)= ln(x) / x. Right side is result that you can get from taking log to both sides of question and arrange for same letter. Such as ln(y)/y=ln(x)/x.
Great explanation, but roots x € R and y € R when x equals y are missed.
Good video. Didn't even think that there could be way of square roots and much more exponents.
you can't take the square root if the number is negative and you can't divide by x because there is no defined domain for x without zero
In the problem the condition that x must be different from y is not set, so, it follows that in the domain of natural numbers there are countless solutions - (x=1, y=1); (x=2, y=2), (x=3, y=3)...
Taking square roots was totally pointless! 😉
y = kx gives x^(kx) = (kx)^x
That is, (x^k)^x = (kx)^x.
→ x^k = kx
→ x^(k-1) = k
→ x = k^(1/(k-1)), y = k^(k/(k-1)).
There was no purpose to extract the square foot. Without it the problem solves much faster
couldn't you just use the x-th root (to the power of 1/x) to get rid of the x-exponents at 0:56 when you did it at 2:43 but with the extra unnecessary steps in-between?
first: to my limited math knowledge (I'm just a chemist not a mathematician), one equation containing two variables always has infinite solutions.
Second: I clearly agree with all the comments below about y=x=0 beeing a solution, as for any y=x would be trivial solution to the problem, as well as obviously the problem of k=1 is totally ignored in the video.
Third: my question: after you solved x=k^(1/(k-1) and y=k^(k/(k-1), why don't you proceed to find a direct term of y=f(x) to at least determine the solution series given for the case you followed previously?
ok, tried it myself, failed directly at searching for algebraic solution for inverse function of k^(1/(k-1))
To be clear, what you're showing is that, for any value of k=/=1, k > 0, there is a pair (x,kx) that satisfy the equation x^kx = (kx)^x. (For k=1, this is true for any positive value of x.) You're not solving an equation so much as you are defining a function.
What happens if you take a glance at it, and immediately realize the equation stands true if x = y = 1?
Or if x=2=y, or x=3=y, or...
if you say, simply, this equation is true if x=y.
2^4=4^2
oh i forgot that
As per assumption y= kx , y directly varies with x. But it is not mentioned in the problem. So why such assumption?
Not an "assumption"!
For _any_ two numbers x and y there is a number k such that y = kx
If k=2 then x=2 and y=4 and its true. If k=1 then x=1^(1/0) and y=1^(1/0) and both of them are indeterminated and equal, so x=y. The solution in video is universal
How can you which is a dependant variable and which is an independent variable. By assuming y=kx u are assuming them to related without any basis. 1 eq’n 2variables so this questions incomplete
Simple. X=1 & Y=1. Done.
But x≠y
I have a very very simple solution to this problem - and can be down without pen and paper. Simple huristics. One ( if not the only ) solution is that x and y are interchangable. so - x=y. anyone want to prove me wrong? Let me add this. You have to have a match between the number of variable and equations to get a real answer. Since we have one equation, there can be no definative answer to this problem. You can get a releationship between x and y, but no definiative answer ( no x = value and no y = value ).
Sure,
X=Y gives an infinite number of solutions,
but it does not give all the solutions.
Good grief!
Ylogx=xlogy
logx/x = logy/y
logx^1/x = logy/y
x^1/x =y^1/y
x=y answer
The solution to this equation is x=1, Y=1
And, x=y, x and y are any positive numbers.
Vers Nice démonstration, but you have to précise that x and y are strictement positive.
This comment section makes me so happy. Nice video, even more awesome responses.
Glad you enjoy it! Thanks 🙏
I need to ask, right where you do the square root, how are you sure that x^kx is positive? Because if it isn't then you can't really do the square root, this means the solutions you will fet from there may or may not be the same as the initial
Taking the square root is unnecessary
every x = y is solving.
1 exp 1 = 1 exp 1
2 exp 2 = 2 exp 2
etc...
Sir could you please next video on this
X^Y ×Y^X=7789 then (X+Y)?
x=exp(-W(-lny/y)) where W is a Lambert function
The problem presented was X to the Y = Y to the X X =? Y = ?. So X can = Y and they can be any non-negative value. K is not needed.
Why would y be proportional to x? Why not assume y = k.ln x?
Or x^2 + y^2 = k or y = k + sin(kx). Maybe it's the only one he could think of?
simply 2^4=4^2
And 5^5 = 5^5.
Cool explanation, but i didnt see the need to sqrt both sides
Lovely algebra, but you can't add a specific equation to the original problem. There is no single answer because there is only one equation for two variables. Thanks for the attempt, anyway!
There are an infinite number of solutions. That's the point!
Hahahah😂😅I dunno what I been told but I've been seeing many murders of crows whilst walking down my roads.
Expones los casos y=0, x=0 y finalmente casos donde x0 e y0, con eso y log sale rápido
Very easy... Answer is x is not equal to y... You now have to find values...😊😂🤣😁😅😉
0:41 why do we even need sqr root here?
It looks pointless. We get rid of it on next steps.
2 to the power of 4 is 16. 4 to the power of 2 is 16. Therefore, x and y = 2 and 4.
X = Y = 1, 2, 3,...etc; Keep It Simple....
Pure mathematics ki yesi taisi kar dali
Unending story !
Все хорошо, но зачем надо было взят корень квадратный?
To solve this
If x=y, > 1^1=1^1; 2^2=2^2....10^10=10^10... 458^458=458^458........
I am not following. Did we need a derivation to show so long as x=y this will be true (save for 0, kind of)?
Ans 1) When x=y
Ans 2)x=2, y=4
Solution - Observation
Ans 3) x=4, y=2
This is not right, why k would be a constant then y and x will have a linear relationship which is assumed and again later we see k is a function of x that means it could not have been a constant, a self contradicting exercise
No, it is just not worded by him properly. He meant that for a given value of x and y (such that x^y=y^x) , k is a constant
X=1, y=1 👍
why did u take square root u could
do it without taking square root
X=y=n
N=1,2,3,4.....
Ничего не понятно, из сказанного. Но в принципе ход мыслей верный.
very fun!
Thanks for the visit
what about when x=y=1,2,3,4,5.... this is a way to get infinite solutions as well and you don't get them by his solution!
Y=kx was not sufficiently motivated.
Why did you start with that assumption?
Not an "assumption".
For _any_ two numbers x and y there's a number k such that y = kx. 😉
Next time : specify the domain of number so that the answer ia ceritain, and practice some more !!
logx(x^y) = y
y * logx(x) = y
logx(x) = 1 so y = y
Зачем извлекать квадратный корень?
x=2, y=4, because 2⁴=4² or 16=16 IT'S TRUE
X = 4
Y = 2
that is what i was thinking.. 1 for all. (no part numbers, no 0., only true numbers, which is 1)
Simple both X and Y equal 3 😂
in here 2:45 you have made a big problem not supposing x≠0
also in here 3:12 you have repeated the same mistake
and here 4:05 not supposing k≠1
I like the way he decides k =3
x=2, y=2 is a solution by inspection
X=1
Y=1
xx=2, y=4
Nice raisonnement
Thanks and welcome 🙏 👍
x=y. any number will be ok.😊😊
No
0?
I follow your explanation up to the point where you said K=3, how do you determine K = 3, this is not explained.
Probably 2^4=4^2 is one answer.
How did you choose number 3?
K = 3 ?
Just "for example..."
I don't think, step at 2:51 is valid. Multiplying the powers with its reciprocal? Which rule is that? I think it's invalid process to solve. Correct me please.
My friend you presented one equation with 2 unknown which is unsolvable, you inserted a second equation out of thin air, which is absolutely no part of the solución, you said y=kx, this is a second equation. I will ask you to solve it using y=sin(px). If you present a problem of this type you need to give both equations at the beginning no after the facts
The assumption of y=kx is wrong as this will restrict the solution of the original equation. In fact, the original equation has infinite solutions because one equation with two unknowns does not have unique solution!!
Can you back up your claim?
Why would it be wrong/solution set restricted when k belongs to the set of real? kx will take every possible real value
@@Axenvyy This is basic vector algebra, one equation two unknowns has infinite solutions!! For example, x = 1 = y, x = 2 = y, etc., all satisfy the equation!
Not an "assumption".
For _any_ two numbers x and y there's a number k such that y = kx. 😉
@@rclrd1This is already an assumption that only two numbers satisfying this condition are true in the subsequent derivation. That means an extra condition is added to qualify the x and y relation!!
The trivial solution is x =1, y = 1. If the question does not rule out that answer, then it is correct. LOL
You are missing the condition that k must not be 1. Because if you put k=1, you are dividing by zero.
x=2, y=4