@@alexyuri_94 i get that u also get an answer but maths is like driving from point B to C in a a road called( A B C) her addingg square root is going from B to A then from A to C .she reached the same destination but wasted petrol
I think you have problems with some concepts. 1) In the change of variable y = k x, k is not a constant but a parameter. 2) You do a lot of unnecessary steps. For example, in the fifth line it is totally unnecessary to extract the square root. 3) In the tenth line you do not analyze what happens if k = 1. In such a case, both sides of the equation cannot be raised to 1/(k-1). In the case k = 1, the parameterization y = k x takes the form y = x, which is the identity function that solves the original equation for all x>0 (first quadrant) and for the values x
Additionally, there is some shortening of expression by dividing of both sides by X. However, there is no analysis what happens if X=0. Same goes for shortening of expression by dividing with K at one point. No analysis what if K = 0.
There are an infinite number of solutions. For every positive value of X, x=y is an obvious solution. Also x=2, y=4 is a solution (as is x-4 and y-2). And so forth....
The solution set is infinite. I think what is described here is taking things around a circle to reach the logic. Way too complicated for a simple solution.
@@French_Horn_Dude the problem with powers with negative base is that to have a real numer, the exponent then needs to be integer or at least a special kind of rational number (or fraction, to put simply) with odd denomiator. Since, for example of (-1)^(-3/2) = 1/sqrt(-1) is problematic (unless you want to use complex numbers), but (-1)^(-5/3) = 1/cuberoot(-1) is kind of fine. Usually, to avoid this, it is assumed that the base is always greater than or equal to zero.
Yes, this was my first reaction also, my guess is that x != y is supposed to part of the question, but it does not get mentioned, hence there are infinite solutions :-)
This is a bit partial as the answer is the solution to two equations, where the linearity of x and y is hidden. But the equation as it is is not bounded by linearity. If you try a different functional form between x and y, you get new answers. And no functional form is required either. Good question though.
There is no reason to take a square root. The key idea is writing y=kx. After this, just a few steps lead to the result x=k^(1/(k-1): Using y=kx in x^y=y^x leads to x^(kx)=(kx)^x = k^x x^x. Divide both sides by x^x. The result is x^(kx-x)=k^x or x^((k-1)x)=k^x. Now raise both sides to the 1/x to get x^(k-1) = k. If k is not equal to 1 then raise both sides to the 1/(k-1) to get x=k^(1/(1-k)). Video then shows that y =kx = k^(k/(1-k)). When k=1, y=x is a solution for any x. One should also stipulate in the statement of the problem that x and y are positive -- otherwise roots of negative numbers might arise.
there is no reason for any of it. there are lots of silly things on the internet. don't take all of them seriously. just examine the silly equation and answer it by inspection. x=y for whatever you want.
How do you decide to substitute y=kx? Is it mentioned in the question? Are we sure that y cannot be a non linear function of x? You just started with the question but it does not help the person trying to learn because they don’t know why you decided to take y=kx and how you reached to that point.
from my experience in olympiads sometimes they expect you to know these tricks thats why some mathematicians are not totally loving the idea of competing in such ways. in other words these problems are not structured at all which is completelly different from the school curriculum, even you take the highest possiblr level into account (a levels further- ib aa-hl). i will say though, that if you see a lot of different exercises similar to this it will probably come to you when you need it if you are good enough that is.
This is kind of a weird question for a test since it has to undergo a lot of analysis for all the trivial vs non trivial solutions. There's not "a solution", single one, so the answer to the exercise would have been much more than just solving for a single value to show proof that "it worked".
I am an amateur who just love to see any math problems that are analyzed and solved. I am also glad to know that there are so many ideas and contributions in comments. The participants reminded me of Martin Gardner in his monthly columns in magazine. Of course, one can only think about it to so much.
The solution is based on a prior guess of a linear relationship y=k x, with k as a constant parameter (parametrization). A more systematic approach, without guessing, is to introduce polar coordinates. Then the same result follows upon finally identifying Tan[theta] with kappa.
As somebody who hasn't done math for 20 years, what you wrote seems conceptually wrong. You have to make make a distinction between x and y because they have a different value. So in principle x is always different from y because you would call the two values the same way
Summarized . . . 5 steps let y = kx and substitute xᵏ = kx, solve for k k = x^(k-1) . . raise both sides to power 1/(1-k) k^1/(k-1) = x . . . sub in equation y = kx y =k^k/(k-1)
I had that same problem a long time ago. There should be a condition y not equal to x. If you differentiate wrt x and equate to zero you will get the value of e. Try it.
I think a better solution is x=1/e^lambert w(-ln(y)/y)because it doesnt involce a random k variable. If you would like to know how to get this feel free to reply to this comment asking.
IMHO the solution is not complete. It was considered only a case for y = kx, need to consider the other cases, probably need to proof that there are no solution other than y = kx The solution can be also a little bit simplified by skipping to take a square root, the following step would be the same, just without 1/2 in power...
As you did, assuming y=kx, there are some problem with your solution that you should mention before...for example, what happen when you consider k=0 and k=1...
Hi Madame, I prefer this solution, 1- between X and Y there is a tangent lets say Y/X = T Now, Y = TX The new equation X ^ TX = XT ^ X If we take the LN of the equation we get this: TX Ln x = X Ln TX Now we reverse again from Ln to the initial formula and we solve it for X X^ T = TX OR X X^(T-1) = TX X ([X^(T-1) - T] = 0 Finaly we get X = 0 ( not a good solution) Or X = e^ [LnT/(T-1)] with T alwasy + and T ]0, 1[ ]1, +inf[ T NOT EQUAL 0 OR 1and not a negative number
Thanks. Your solution gives a more generalised solution on top of the trivial solution x=y. What can we do if the question is to find the full solution to the problem ? For example, non-linear relation between x &y, or even extend the solution to complex plane.
furthermore, looking at other cases (negative x, y, x^y and/or y^x) countable pairs (interesting!) can be found using the same logic. the squirreliness of X^Y in the face of unrestricted real arguments makes for headaches solving and bigger ones reading the solution, but for instance it is an easy EFS that for x=2/3 there exists a y in (-1,0) such that X^Y = Y^X.
yeah it's the same thing I thought. It's only changing the power that is then later eliminated. Instead of eliminating x/2 by doing the power 2/x, one could just do that with 1/x.
1. Why do you take square roots initially ? 2. What allows you to take as an assumption that y=Kx ? Why should the ensemble of solutions be limited to this special case ?
1. The square roots step is redundant and serves no purpose. 2. The assumption y=Kx holds true as long as x ≠ 0, which is anyway necessary here (just substitute x = 0 in the original equation, then you get y = 0 as well which makes it 0^0 indeterminate form). So y=kx is not really an assumption but a parametrization. We are simply calling the ratio (y/x) as "k" and representing the solution space in this form as it has a much simpler structure.
Please, correct me if I'm wrong somewhere... We have one equation with x and y. What we can do is to find the function in its explicit form y(x) that satisfies this equation. There's no way how we can find any numerical values, because for that we would need 2 equations!! Here's my solution: 1. Logarithmize of both sides: y * ln(x) = x * ln(y) (here actually any log of any positive base "a" would work) 2. ln(x) / x = ln(y) / y 3. Exponentiate both sides: exp (ln(x)/x) = exp(ln(y) / y). (here I used "e" as a base, but like in Step1 same arbitrary positive "a" as the base would also work) 4. exp(ln(x))^(1/x) = exp(ln(y))^(1/y) 5. x^(1/x) = y^(1/y). 6. Compare LHS and RHS from Step5 and see that the only option for this to hold is y=x. So, y=x is the only solution. So, whatever values you put as "x", if you put same for "y", the equation will hold. In your case, for y = kx, this means that k can be only 1. In special case, using a bit of limits and calculus, it can be shown that same holds also for x = 0 and y =0. UPD: I just realized the following.. If we have expression like f(x) = f(y), this leads to x=y only if the function f is such that a certain value of f(x) corresponds to only one x, like for example if it is monotonic for the whole interval of x of interest. Then, the inverse function f^-1(x) is well-defined. In our case a function f(x) = x^(1/x) is not such a function. As f is below 1, there is only one x that corresponds f(x), however in between [1 , e^(1/e)] there are two x values that are mapped to the same function value. You can see, for example, that 1^1 = 1, so that f(1) =1, but at the same time when x is infinetely large (x->+infinity), f also tends to 1 (x^(1/x)=exp(lnx/x)->exp(0)=1). These two branches coalesce at the point (e, 1^(1/e)). For any x
5:52: "let's find out y". That's what I was trying to do the whole time K was tossed in there. >>;=) Obviously, this video wouldn't exist if what you were doing didn't work, but I was skeptical about adding a 3rd letter in there (even if it was a constant). Kind of bizarre, but an interesting way to solve this. Your primary decisions to make y=kx and to √ both sides is something I was trying to determine why you did it that way. Everything thereafter made sense. Thanks for sharing!
Clearly, we must find the kernel of the transform in R3 to imbed this R2 equation to see it in all of its glorious forms Informing ourselves of this higher dimensional vision of this projection we can see it clearly from all angles. Now to do this most young people should take a gummy and realize the whole thing is just a prank. x=y, no math no bother, just a way to play with your head.
I attempted the question differently. From some trial solutions like x = y and (2, 4), I saw a pattern like y = x^n which implies that y and x should have the same base in the general solution. Therefore, I assumed x = b^m for m > 0 and any b > 0. My solution is ( [1+1/m]^m, [1 + 1/m]^(m+1)). For example, if you put m = 4, LHS x^y = [5/4]^(5^5/4^4) = y ^x.
But when you set y=kx, you have established that you are looking for solutions where y is linear dependent of x and that has to be proven or assumed looking ONLY for such subset of solutions.
As k is constant, the problem has infinite solutions where k >= 0. Besides, by introducing an assumption that y is linearly related to x proves that the original problem cannot be resolved without stating additional condition to close the system..
how about the combi 0;0 and 1;1 so basically x=y ? One thing wasn't mentioned when discarding the 1/x and x/1 that it is only allowed when x0 but as it is one of the solutions you can't ignore it
Think about what is in front of you for maybe a minute and it might cross your mind that 2^4 = 16 and 4^2 = 16. x = 2 and y = 4. Not all that many numbers are a product of multiple exponential expressions.
just want to point out when you assume the exponential is not integers, you lose x,y as negative integers. The solution is you missed is all negative integers, x=y= -1, -2, -3, etc. I am not clear whether you are aware of the obvious solution for positive real numbers when k=1 as well.
Your conclusion is correct, if one let's k equal 1. Unless one follows the speakers logic to the end. Then k cannot equal 1. Hence, y cannot equal x to satisfy the speaker's final solution. So, are two branches to the solution?
What exactly is the question at hand? To introduce a constant (parameter) then change the parameter to find the corresponding x and y values is not implied in the question.
It is a strange method that missed many solutions. For example it misses the solution: x = 2, y = 2k, where k is the solution of the equation k = -2^(k-1).
condition is obvious x = y, the magical thing here is that for the mind the conclussion is instantly simple, but the logical steps to prove your mind is right (self-test) are way more complicated...
Am biased by my profession as engineer and tried solving it brute force by observation. 2 raised to 4 is 4 raised to 2. Been 30 ears since i solved equations. Look to visualize and seek approximations than resort to rigorous proofs. Enjoyed looking at solution
Thank you for the solution. Putting aside trivial solutions/edge cases, I was so confused by the statement that k is a constant from the very beginning… I guess it’s better to say that for any (x,y) that are solutions of the equation above there is a k that fits y=kx, so let’s go from there
I have an additional solution! x = y = C, where C is a constant different from 0. For example with C = 3, then we have 3^3 = 3^3, haha. It's the same as setting k to 1 in your equation y = k*x.
Yes, that was an apparent (?) solution, but it is interesting that it doesn't fall out of her result as you can't set k=1 using it. I haven't taken the time to figure out why, it would be interesting to see why it doesn't though.
None of the trivial x=y solutions are accounted for in the final formula so they should have been noted separately. The question of negative x and y values were not answered anywhere either. The main y=kx idea is correct but the relaxed use of roots and powers without checking if they can be applied and accounting for possible lost solutions alone shows that the implementation of the idea was overall quite sloppy. My old math teachers would not be proud.
@@Kirillissimusthere’s also where x=2 and y=4 (or the other way around). 2^4=4^2. In her equation, k cannot equal 1 as a value of 1 results in a divide by 0 in the equation, since the denominator of the exponent is k-1.
Devinding by x and k-1 without any conditions is a lil bit confusing for me, how about if we put k=1, it s gonna give a problem, and how about if x=0, and u devide by 0 in that step, and the second problem the infinite solutions when x = y, maybe the problem is resolved partially and need more steps to fully solve it .
Actually the answer is x=y, x is any value, or x=k^{1/(k-1)} and y=k^{k/(k-1)}, where k is any value greater than 1. For example, k=2, x=2, y=4, or k=5, x=4th root of 5, y=4th root of 5^5.
Your answer considering k=3 and finding answer of x=sqrt(3) is not satisfying y=kx as you found answer of y as [sqrt(3)]^3 which should be 3*[sqrt(3)] as per your assumption of y=kx Answer to yout question can be x=1 and y=1 so to satisfy all assumptions and equations.
By looking at it I thought x = y for all real numbers. What other solution could there be? Then I watched the video and agree with the rest of the solution.
The solution is just x = y. But I miss the definition of x and y element from what? How to get it. first multiply the power of both sides with 1/y this gives you x^1 = y^(x/y) now multiply the power of both sides with 1/x this gives you x^(1/x) = y^(1/y) There is a law a^(1/a) = b^(1/b) then a = b in our case x = y
Why do we take, y=kx, we should substitute any of the variables by other variables, which are not used in the given problem. Because all" x" mix together.
If x=k^1/k-1 then k=x^k-1 by same way you find that k=y^(k-1)/k X^k-1=y^(k-1)/k Y=x^k Which mean that for every value of k there are infinite solutions for x and y not only one As you assumed if k=3 then y=x^3 not only 2 values for x and y
Not sure why i watched the entire video when i understand 0% of what being said. Please someone tell me im not the only one who doesnt know whats happening 😢
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I don't understand why was necessary to square root when it would have been easier to rise direct to power (1/x) instead of (2/x)
This is the problem with byhearting the math solution rather than using logic.
It was not necessary, but a solution is still a solution. We can reach the same destination by different paths, and that's the beauty of it
I was wondering the same think bro
@@alexyuri_94 i get that u also get an answer but maths is like driving from point B to C in a a road called( A B C) her addingg square root is going from B to A then from A to C .she reached the same destination but wasted petrol
You are right. SQRT(2) is useless.
I think you have problems with some concepts.
1) In the change of variable y = k x, k is not a constant but a parameter.
2) You do a lot of unnecessary steps. For example, in the fifth line it is totally unnecessary to extract the square root.
3) In the tenth line you do not analyze what happens if k = 1. In such a case, both sides of the equation cannot be raised to 1/(k-1).
In the case k = 1, the parameterization y = k x takes the form y = x, which is the identity function that solves the original equation for all x>0 (first quadrant) and for the values x
Perfect. Absolutely correct
Why y=Kx?
@@HenriqueSantos-xd1egjust because😂
Additionally, there is some shortening of expression by dividing of both sides by X. However, there is no analysis what happens if X=0. Same goes for shortening of expression by dividing with K at one point. No analysis what if K = 0.
y=kx , k = constant , that mean y Directly proportional to x therefor no problem
There are an infinite number of solutions. For every positive value of X, x=y is an obvious solution. Also x=2, y=4 is a solution (as is x-4 and y-2). And so forth....
So the problem is not solved by she?
Hahaha
The answer is a graphics y=x
X=2, y=4 is her solution for k=2.
The solution set is infinite. I think what is described here is taking things around a circle to reach the logic. Way too complicated for a simple solution.
x = 1, y = 1 can also be a solution (if there is no other constraints such as x != y)
That was my first thought. x=1, y = 1, then x = 0 and y = 0 and basically x = y
Seems there are infinite solutions where x = y, since that condition was not excluded.
That's what I say
x=y
@@French_Horn_Dude the problem with powers with negative base is that to have a real numer, the exponent then needs to be integer or at least a special kind of rational number (or fraction, to put simply) with odd denomiator. Since, for example of (-1)^(-3/2) = 1/sqrt(-1) is problematic (unless you want to use complex numbers), but (-1)^(-5/3) = 1/cuberoot(-1) is kind of fine. Usually, to avoid this, it is assumed that the base is always greater than or equal to zero.
Yes, that seemed obvious, x=y.
Why not just answer that due to symmetry x = y?
@@hlindstrombecause there are also infinite solutions where x≠y
As long as x=y, there are infinite number of roots. 👍👍👍
Yes, this was my first reaction also, my guess is that x != y is supposed to part of the question, but it does not get mentioned, hence there are infinite solutions :-)
The vital part of the question should be y != x. Otherwise, the simple answer is y = x = 1
x=2, y=4@@chrismcgowan3938
@ajaysamanta9661 😶🌫️
And ironically she did not mention that k may not be equal 1 in their solution
This is a bit partial as the answer is the solution to two equations, where the linearity of x and y is hidden. But the equation as it is is not bounded by linearity. If you try a different functional form between x and y, you get new answers. And no functional form is required either. Good question though.
Off course, three more answers are possible are (x = 1, y = 1), (x = 2, y = 2), (when x = 2, y = 4, when x = 4, y = 2).
@@nikkverma5523This is what I thought no need to do anything
@@nikkverma5523 (x,x) in general is a solution.
Dude, it was a prank. See my comment above about dihydrogen monoxide
There is no reason to take a square root. The key idea is writing y=kx. After this, just a few steps lead to the result x=k^(1/(k-1): Using y=kx in x^y=y^x leads to x^(kx)=(kx)^x = k^x x^x. Divide both sides by x^x. The result is x^(kx-x)=k^x or x^((k-1)x)=k^x. Now raise both sides to the 1/x to get x^(k-1) = k. If k is not equal to 1 then raise both sides to the 1/(k-1) to get x=k^(1/(1-k)). Video then shows that y =kx = k^(k/(1-k)). When k=1, y=x is a solution for any x. One should also stipulate in the statement of the problem that x and y are positive -- otherwise roots of negative numbers might arise.
x= k^(1/(k-1) and y = k^( k/k-1) when k=constant ,are the solutions of equation and the next steps are for checking
When k=1 => the solution is x=y=e. That assuming y=k * x. In general, when x >0 and x
there is no reason for any of it. there are lots of silly things on the internet. don't take all of them seriously. just examine the silly equation and answer it by inspection. x=y for whatever you want.
@@weeblyploonbottom810 Silly "things" on internet for sure.... 😁😁
Why assuming y=kx? That only gives a family of solutions
Correct. If you don't assume y=kx, then you need to use the Lambert-W function which would give you other solutions as well.
If x!=0, then you can always define k=x/y.
And then you can look for solutions corresponding to each value of k.
Given any x, can't you get any value of y by substituting the correct k value? In other words, wouldn't this form provide all the solutions?
How do you decide to substitute y=kx? Is it mentioned in the question? Are we sure that y cannot be a non linear function of x?
You just started with the question but it does not help the person trying to learn because they don’t know why you decided to take y=kx and how you reached to that point.
from my experience in olympiads sometimes they expect you to know these tricks thats why some mathematicians are not totally loving the idea of competing in such ways. in other words these problems are not structured at all which is completelly different from the school curriculum, even you take the highest possiblr level into account (a levels further- ib aa-hl). i will say though, that if you see a lot of different exercises similar to this it will probably come to you when you need it if you are good enough that is.
This is kind of a weird question for a test since it has to undergo a lot of analysis for all the trivial vs non trivial solutions. There's not "a solution", single one, so the answer to the exercise would have been much more than just solving for a single value to show proof that "it worked".
I am an amateur who just love to see any math problems that are analyzed and solved. I am also glad to know that there are so many ideas and contributions in comments. The participants reminded me of Martin Gardner in his monthly columns in magazine. Of course, one can only think about it to so much.
The solution is based on a prior guess of a linear relationship y=k x, with
k as a constant parameter (parametrization). A more systematic approach, without guessing, is to introduce polar coordinates. Then the same result follows upon finally identifying Tan[theta] with kappa.
Logically, if x=y, then any number satisfies the equation
As somebody who hasn't done math for 20 years, what you wrote seems conceptually wrong. You have to make make a distinction between x and y because they have a different value. So in principle x is always different from y because you would call the two values the same way
Where are you trying to find the solutions? In C, R, Q, Z or N?
For example all the points on the line y = x in the first quadrant are also solutions.
Except 0, as 0^0 is undefined
Apologize for my poor math skills, but doesn’t this have infinite solutions (x=y)?
look at my reply above. it was a joke problem. you may find my silly answer as funny as the problem. Of course x=y for anything
Summarized . . . 5 steps
let y = kx and substitute
xᵏ = kx, solve for k
k = x^(k-1) . . raise both sides to power 1/(1-k)
k^1/(k-1) = x . . . sub in equation y = kx
y =k^k/(k-1)
is just an equation with two unknowns. It has infinite solutions. I don't understand why they are supposed (or chosen) to be on a line
Dad carries his son on his shoulder and when the dad becomes older, the son carries his dad on his shoulder: PROVED
Lmao
I had that same problem a long time ago. There should be a condition y not equal to x. If you differentiate wrt x and equate to zero you will get the value of e. Try it.
I think a better solution is x=1/e^lambert w(-ln(y)/y)because it doesnt involce a random k variable. If you would like to know how to get this feel free to reply to this comment asking.
IMHO the solution is not complete.
It was considered only a case for y = kx, need to consider the other cases, probably need to proof that there are no solution other than y = kx
The solution can be also a little bit simplified by skipping to take a square root, the following step would be the same, just without 1/2 in power...
As you did, assuming y=kx, there are some problem with your solution that you should mention before...for example, what happen when you consider k=0 and k=1...
Along with the case where x=0, before using the 2/x simplification even though it's obvious
when saying y=kx and k=constant that mean k>0 ,I think the question is closer to physics than mathematics
Hi Madame,
I prefer this solution,
1- between X and Y there is a tangent lets say
Y/X = T
Now, Y = TX
The new equation
X ^ TX = XT ^ X
If we take the LN of the equation we get this:
TX Ln x = X Ln TX
Now we reverse again from Ln to the initial formula and we solve it for X
X^ T = TX OR
X X^(T-1) = TX
X ([X^(T-1) - T] = 0
Finaly we get
X = 0 ( not a good solution)
Or
X = e^ [LnT/(T-1)] with
T alwasy +
and T ]0, 1[ ]1, +inf[
T NOT EQUAL 0 OR 1and not a negative number
I thought u where coming good until ur 1st new equation, that right side doesn't add up I think
Qué pasaba si en vez de elegir K=3 se elegía otro valor?
Thanks.
Your solution gives a more generalised solution on top of the trivial solution x=y.
What can we do if the question is to find the full solution to the problem ? For example, non-linear relation between x &y, or even extend the solution to complex plane.
f(u) = logu/u on u>1 has two continuous monotone branches {(1
furthermore, looking at other cases (negative x, y, x^y and/or y^x) countable pairs (interesting!) can be found using the same logic. the squirreliness of X^Y in the face of unrestricted real arguments makes for headaches solving and bigger ones reading the solution, but for instance it is an easy EFS that for x=2/3 there exists a y in (-1,0) such that X^Y = Y^X.
Why to take square root?
One get same result without taking square root.
That was absolutely bizarre. There is nothing in the original equation to even suggest taking square root. There was no two in the exponents.
yeah it's the same thing I thought. It's only changing the power that is then later eliminated. Instead of eliminating x/2 by doing the power 2/x, one could just do that with 1/x.
Exactly my thought! Taking the square root makes no sense.
A cube root or any other power other than zero would make it look even more "smart" albeit completely unnecessary!
1. Why do you take square roots initially ?
2. What allows you to take as an assumption that y=Kx ? Why should the ensemble of solutions be limited to this special case ?
Same question
1. The square roots step is redundant and serves no purpose.
2. The assumption y=Kx holds true as long as x ≠ 0, which is anyway necessary here (just substitute x = 0 in the original equation, then you get y = 0 as well which makes it 0^0 indeterminate form). So y=kx is not really an assumption but a parametrization. We are simply calling the ratio (y/x) as "k" and representing the solution space in this form as it has a much simpler structure.
What about x=y=1, x=y=2 and x=2,y=4?
the first square-root seamed unnessesery,
Indeed
Please, correct me if I'm wrong somewhere... We have one equation with x and y. What we can do is to find the function in its explicit form y(x) that satisfies this equation. There's no way how we can find any numerical values, because for that we would need 2 equations!! Here's my solution:
1. Logarithmize of both sides: y * ln(x) = x * ln(y) (here actually any log of any positive base "a" would work)
2. ln(x) / x = ln(y) / y
3. Exponentiate both sides: exp (ln(x)/x) = exp(ln(y) / y). (here I used "e" as a base, but like in Step1 same arbitrary positive "a" as the base would also work)
4. exp(ln(x))^(1/x) = exp(ln(y))^(1/y)
5. x^(1/x) = y^(1/y).
6. Compare LHS and RHS from Step5 and see that the only option for this to hold is y=x.
So, y=x is the only solution. So, whatever values you put as "x", if you put same for "y", the equation will hold. In your case, for y = kx, this means that k can be only 1.
In special case, using a bit of limits and calculus, it can be shown that same holds also for x = 0 and y =0.
UPD: I just realized the following.. If we have expression like f(x) = f(y), this leads to x=y only if the function f is such that a certain value of f(x) corresponds to only one x, like for example if it is monotonic for the whole interval of x of interest. Then, the inverse function f^-1(x) is well-defined. In our case a function f(x) = x^(1/x) is not such a function. As f is below 1, there is only one x that corresponds f(x), however in between [1 , e^(1/e)] there are two x values that are mapped to the same function value. You can see, for example, that 1^1 = 1, so that f(1) =1, but at the same time when x is infinetely large (x->+infinity), f also tends to 1 (x^(1/x)=exp(lnx/x)->exp(0)=1). These two branches coalesce at the point (e, 1^(1/e)). For any x
5:52: "let's find out y". That's what I was trying to do the whole time K was tossed in there. >>;=) Obviously, this video wouldn't exist if what you were doing didn't work, but I was skeptical about adding a 3rd letter in there (even if it was a constant). Kind of bizarre, but an interesting way to solve this.
Your primary decisions to make y=kx and to √ both sides is something I was trying to determine why you did it that way. Everything thereafter made sense. Thanks for sharing!
Clearly, we must find the kernel of the transform in R3 to imbed this R2 equation to see it in all of its glorious forms Informing ourselves of this higher dimensional vision of this projection we can see it clearly from all angles. Now to do this most young people should take a gummy and realize the whole thing is just a prank. x=y, no math no bother, just a way to play with your head.
@@weeblyploonbottom810 I feel like the matrix just had an uh oh. >>:=p
5) if x is even, the step of power to 2/x result in a modular equation.
I find your videos fascinating. I don't know why I keep watching it before I sleep though.
I attempted the question differently. From some trial solutions like x = y and (2, 4), I saw a pattern like y = x^n which implies that y and x should have the same base in the general solution. Therefore, I assumed x = b^m for m > 0 and any b > 0.
My solution is ( [1+1/m]^m, [1 + 1/m]^(m+1)). For example, if you put m = 4, LHS x^y = [5/4]^(5^5/4^4) = y ^x.
But when you set y=kx, you have established that you are looking for solutions where y is linear dependent of x and that has to be proven or assumed looking ONLY for such subset of solutions.
How you give the power 2/x at both side when both side base is different
What was the step of making the square root for?
Just complicating things or just making the video longer!??😂😂😂
Its a single equation with 2 variables. There are an infinite number of x,y values
As k is constant, the problem has infinite solutions where k >= 0. Besides, by introducing an assumption that y is linearly related to x proves that the original problem cannot be resolved without stating additional condition to close the system..
Any two real numbers are linearly dependent. dim R = 1.
how about the combi 0;0 and 1;1 so basically x=y ? One thing wasn't mentioned when discarding the 1/x and x/1 that it is only allowed when x0 but as it is one of the solutions you can't ignore it
Wow such a good brain you have there! ❤🎉
You also have the trivial case where x=y (as long as x, y not equal to zero)
quick question, do x and y have to be different? if no why dont we just take x = 1, y = 1
Correct me if I'm wrong but, if x = y won't that work also?
Think about what is in front of you for maybe a minute and it might cross your mind that 2^4 = 16 and 4^2 = 16. x = 2 and y = 4. Not all that many numbers are a product of multiple exponential expressions.
just want to point out when you assume the exponential is not integers, you lose x,y as negative integers. The solution is you missed is all negative integers, x=y= -1, -2, -3, etc. I am not clear whether you are aware of the obvious solution for positive real numbers when k=1 as well.
it is easy to found the y=x is one of group value,so x and y is able to every no.
Can you explain why you have taken y=kx. How can you say that yis always a multiple of x. It's true that y=kx ... Reason also should be known
Who says that this should be a linear relationship between y and x?
yeah, i agree with you..
This should be true for every value where x=y
And this is in total contradiction with the proposed solution where k=1 is excluded
Yes, x=y is a solution only that x and y not equal to zero
Your conclusion is correct, if one let's k equal 1. Unless one follows the speakers logic to the end. Then k cannot equal 1. Hence, y cannot equal x to satisfy the speaker's final solution. So, are two branches to the solution?
Isn't this equation satisfied for all X=Y? Maybe excluding zero? Is 0 to the power 0 even defined?
For k= 1, what is the value of x and y
Hello and Thanks. I do not understand how do you choose "...If k = 3, ..." at 5:18 . Can you please explain it
No meaning.
This problem could have been solved in 4 steps by applying the rule of logarithm, etc.
The ultimate proof being x/y = 1 or x= y.
Excellent
También fácilmente le puedes asignar el valor de 1 a ambas variables y listo.
What exactly is the question at hand? To introduce a constant (parameter) then change the parameter to find the corresponding x and y values is not implied in the question.
The simplest solution is where x=y for that condition any given number is correct.
Ok, idk how to do that, but if both is equal to 1, its correct?
Chiarissima dimostrazione, veramente eccellente.....grazie !♥️
It is a strange method that missed many solutions. For example it misses the solution: x = 2, y = 2k, where k is the solution of the equation k = -2^(k-1).
But k cannot be 0. So all solutions where x = y are missed, though y = kx is valid when k=1
x=1 y=1 or in short x=1=y no calculation needed. Valid solution?
how about if the first step,we divide by X^Y at the both side
What about 3 when both X and Y are value 3, just because different variables, does not mean cannot have same value
You transfomed the original equation
condition is obvious x = y, the magical thing here is that for the mind the conclussion is instantly simple, but the logical steps to prove your mind is right (self-test) are way more complicated...
Does it work if x = 17? or is this just a trick?
It was going fine until u took k = 3 so randomly.
Am biased by my profession as engineer and tried solving it brute force by observation. 2 raised to 4 is 4 raised to 2. Been 30 ears since i solved equations. Look to visualize and seek approximations than resort to rigorous proofs. Enjoyed looking at solution
As there isn't any restriction in command, then the answer is x=1 and y=1.
Thank you for the solution. Putting aside trivial solutions/edge cases, I was so confused by the statement that k is a constant from the very beginning… I guess it’s better to say that for any (x,y) that are solutions of the equation above there is a k that fits y=kx, so let’s go from there
when ppl say constant in engineering they often just mean it doesn't depend on other variables here, at least that's how I interpreted it
This is right for every x=y case apart from 0
I have an additional solution! x = y = C, where C is a constant different from 0. For example with C = 3, then we have 3^3 = 3^3, haha. It's the same as setting k to 1 in your equation y = k*x.
Yes, that was an apparent (?) solution, but it is interesting that it doesn't fall out of her result as you can't set k=1 using it. I haven't taken the time to figure out why, it would be interesting to see why it doesn't though.
@@SpacePhys At 3:30 you can't divide both sides by k-1, without implicitly assuming k ≠ 1.
Very Beautiful Question !!!!!!
👌🏾👍🏾🤙🏾
👏👏👏👏👏
I see two solutions by inspection: x=y, and either of them equals 2 and the other 4.
How about trivial solutions like x=y=1 or x=y=2?
None of the trivial x=y solutions are accounted for in the final formula so they should have been noted separately. The question of negative x and y values were not answered anywhere either. The main y=kx idea is correct but the relaxed use of roots and powers without checking if they can be applied and accounting for possible lost solutions alone shows that the implementation of the idea was overall quite sloppy. My old math teachers would not be proud.
@@Kirillissimusthere’s also where x=2 and y=4 (or the other way around).
2^4=4^2.
In her equation, k cannot equal 1 as a value of 1 results in a divide by 0 in the equation, since the denominator of the exponent is k-1.
Devinding by x and k-1 without any conditions is a lil bit confusing for me, how about if we put k=1, it s gonna give a problem, and how about if x=0, and u devide by 0 in that step, and the second problem the infinite solutions when x = y, maybe the problem is resolved partially and need more steps to fully solve it .
I never understood math when you simply added some letters and then eliminated them later
There are a set of values for x and the corresponding values of y for all real integers..
Why aren’t all ordered pairs x=y from negative infinity to positive infinity solutions?
Thank you Mom for your great videos. Could you please give me the link for these questions references?
Actually the answer is x=y, x is any value, or x=k^{1/(k-1)} and y=k^{k/(k-1)}, where k is any value greater than 1. For example, k=2, x=2, y=4, or k=5, x=4th root of 5, y=4th root of 5^5.
Your answer considering k=3 and finding answer of x=sqrt(3) is not satisfying y=kx as you found answer of y as [sqrt(3)]^3 which should be 3*[sqrt(3)] as per your assumption of y=kx
Answer to yout question can be x=1 and y=1 so to satisfy all assumptions and equations.
Can i simply say 2 and 4? With no calculation but the equation is correct
👍👍amazing....
What a long trip around the barn!
3:24 I realised just now that u can multiply powers like that too 😮
this question goes against the basic principle that the number of equations has to be equal to the number of unknown variables
the solution can also be x=1 and y=1. X to the power 1 = 1; Y to the power 1=1.
Eu desconbri essa fórmula em 2005 quando tinha 18 anos.
By looking at it I thought x = y for all real numbers. What other solution could there be? Then I watched the video and agree with the rest of the solution.
negative integers also work if x=y
@@dongxuli9682 and (I suspect) complex numbers.
@@fluffyduck1944 Can't see why it wouldn't work with complex numbers. If X = Y then X^Y = Y^X = X^X = Y^Y, I think.
The solution is just x = y. But I miss the definition of x and y element from what?
How to get it.
first multiply the power of both sides with 1/y this gives you
x^1 = y^(x/y)
now multiply the power of both sides with 1/x this gives you
x^(1/x) = y^(1/y)
There is a law
a^(1/a) = b^(1/b) then a = b in our case x = y
Why do we take, y=kx, we should substitute any of the variables by other variables, which are not used in the given problem. Because all" x" mix together.
If x=k^1/k-1 then k=x^k-1 by same way you find that k=y^(k-1)/k
X^k-1=y^(k-1)/k
Y=x^k
Which mean that for every value of k there are infinite solutions for x and y not only one
As you assumed if k=3 then y=x^3 not only 2 values for x and y
I am not a mathematician, but "lets assume k=3" and assume anything for all the other numbers seems dumb to me.
Also check the trivial solution x=y=1
Not sure why i watched the entire video when i understand 0% of what being said. Please someone tell me im not the only one who doesnt know whats happening 😢