Russian Maths Olympiad Question

[@ChavoMysterio]

By observation... x=2, y=4 x=4, y=2

[@user-ry2yw5lf2r]

There are an infinite number of solutions. These solutions must obey the condition y = x^k. By applying this condition the equation x^y = y^x becomes x^x^k = x^k^x, that is x^kx = x^kx, which is true.

[@AlainPaulikevitch]

How do you translate what you said into an equation that gives y based on values of x? is that even possible? Using the formula from the video for x and y based on k with k > 0 and k 1 allows finding x and y while k varies. However the y = k.x form which allows this, does work but is not explained/justified while your form seems more appropriate, however it leads to a tautlogy. Deimos on android plots x^y = y^x and it seems like it should have a clean formula that i can't figure out

[@Kokujou5]

guys... am i the only one that noticed that literally infinite solution exists? like every constellation of numbers is valid as long as y = x 2^2=2^2, 3^3=3^3, 4^4=4^4

[@Tommy_007]

No, you are not the only one... It is in the video!

[@geraldsmith6225]

That's too easy

[@user-lj1nd8rq9w]

WTF I just saw? This one has: 1. Indefinite number of trivial solution where x=y 2. Two non-trivial integer solution x=2; y=4 and x=4; y=2 3. Indefinite number of real solution like x=3; y=2.47805268 or x=2.2; y=3.4776896 and so on. KZbin became special place where people who have no clue in subject trying to tech everyone else in this very subject. And it looks like all authors of those "mathematical" clips just learned to write and proudly show to whole world new acquired skills. Common solution. x=a^(1/(a-1)) and y=ax where a any number>0. It give us x=2, y=4 when a=2 and x=4, y=2 when a=0.5. Also there is very interesting result x=2.25, y=3.375 when a=1.5 And plethora of solutions anywhere in between.

[@mohamedhamouch1122]

Hi; There is a typo error at 12:05. kx in the right side of the equation should be closed in parenthesis but the next is good.

[@fredsmith6324]

you mean at 2:23? yeah, she corrected it at 3:05

[@ShrikrishnaMarathe]

i dont know if y=kx is a valid assumption. so it may only give partial solutions?

[@sunilchaudhari970]

(2,4) or (4,2) are the solutions by observation😂

[@aruntiwari5714]

Good 😊

[@abuabu2441]

X=y😂😂😂 X=1000 Y=1000 X^1000=y^1000 😂😂😂😂

[@softlinkmangalore4133]

X and Y are two different variables. Two variables cannot have the same values.

[@lnmukund6152]

When x=Ky ,thesolutions are always x=k^1/k-1, y=k^k/k-1, Mukund

[@manojchaugule794]

very nice🙏

[@alikandis2549]

I have to admit it is brilliant letting k be your independent variable for one aspect, you can generate infinite solutions, also includes x=y solution. But if your aim is to assign a value for one of the variables, you cannot calculate the other. For example, at one of your infinite solutions, x=5. We know but cannot calculate y if x=5.

[@vaclavkrpec2879]

What was the square root for, then? :-)

[@arvindkulkarni6580]

very nice solution mam

[@ericfleet9602]

Ummm, this is silly... any solution where x = y is going to be correct. There is no unique solution to this problem as presented.

[@Tommy_007]

Only children in primary school assume that an equation must have a unique solution...

[@Tommy_007]

The solution in the video generates infinitely many solutions.

[@ericfleet9602]

@@Tommy_007 Cute, but stupid response. It is supposedly a Math Olympiad Question, meaning they would not give a question that is so trivial. But, only a child wouldn't realize that.

[@Observer1973]

May you prove there are no other solutions except y=x?

[@SylvainDemuyter]

​@@Observer1973Yes by a simple analysis of x/log(x) function.

[@entertainmentzone6838]

X=1, Y=1 will also satisfy this inequality

[@user-go5sy7mu6q]

Nice

[@LaxmanKumar-yi4cv]

Your video is great 👍

[@OneEyedJacker]

I got ln(x)/sqrt(2) = ln(y)/sqrt(2) after integrating both sides, which simplifies to x=y

[@vraser]

Huh?? Seriously???? The answer to this is (any number, where x=y)

[@stevemaskal5933]

no0t all negative numbers

[@marklevin3236]

x=y. ,as long as x^x is defined.....x is either a positive number or integer negative are the solutions

[@rolandkaschek9722]

You have assumed that x = k* y, for a constant k. That will make you miss out on many solutions! After all, as others have pointed out, (x,y)=(r,r) is a solution for each reall number r.

[@Tommy_007]

No. Choose k=1...

[@rolandkaschek9722]

@@Tommy_007 thank you. Indeed I overlooked the obvious solution.

[@harshnambiar]

Yeah but you'd still lose all the higher power solutions like for x = ky2

[@bozhidarpenkov7972]

How about 2^4=4^2 ???

[@conigedegato2681]

I thought the same!!!!

[@Tat52]

k=2, then x=2, y=kx=2*2=4

[@Tat52]

But that equation has many solutions!

[@lashaghetia8709]

Y=k×k^1/k-1

[@valeryshebaldenkov9326]

И где тут russian? Решение звиздоватое... И абсолютно неверное.

[@arvindkulkarni6580]

you are really a good teacher for math Olympiad problem

[@user-wi8iq3hn3k]

У нас за такое решение ставят 0 баллов из 7. Не нужно браться за задания, которые намного выше вашего арсенала.

[@user-lj1nd8rq9w]

Моя училка за такое решеное забила-бы меня учебником по алгебре :-)))))

[@tapaskantimitra2881]

How you are assuming k is not equal to 1?

[@sudhirsharma7966]

Nobody can solve 2 variables by one equation 😮

[@user-lj1nd8rq9w]

It is not always true. For example x+y+z=30; x^2+y^2+z^2=300 has only one solution and it is possible to show it positively. However in this particular case there are infinite number of solutions.

[@uduehdjztyfjrdjciv2160]

x=y trivial solution and infinitely many solutions in lambert function x=y=e is both.

[@WhiteGandalfs]

(1:20) The square root is redundant.

[@herbertklumpp2969]

Trivial, x=y is always a solution x,y >0

[@abuabu2441]

0^0=0^0=>1=1

[@herbertklumpp2969]

@@abuabu2441 only if 0^0 is well defined

[@shubhrodutta]

What about x=2 and Y = 4

[@Tommy_007]

Good video. Unfortunately, many viewers don't understand it... They don't understand that k is a parameter that generates the (infinitely many) solutions.

[@Mamtamaam]

Yes, you are right

[@jayktomaszewski8738]

what about 2^4 = 4^2

[@pedrofajardo8137]

....... Porque? Y = kx..... Nadie lo dice en el problema.

[@dr.walterstadler1837]

Infinite number of solutions for x = y

[@LaxmanKumar-yi4cv]

Hello 👋

[@LaxmanKumar-yi4cv]

How are you ☺

[@RichardSalvucci]

trivial

[@humelakecabin]

I got X =4 and Y =2 16

[@ShareTyro2010]

x^y= y^x , ln(x^y )=ln(y^x ), y.ln(x)=x ln(y) y/x=ln(y)/ln(x) , y/x=log_x(y) let k= y/x where k is constant obviously for k=1 it follows x=y ≠0, k =log_x(kx) k=log_x(x)-log_x(k)=1-log_x(k) k-1= log_xk =>x^(k-1)=k or x= k^(1/(k-1)), y=k^(k/(k-1)) check for k=2 x= 2^(1/(2-1))=2, y=2^(2/(2-1))=2^2=4, 2^4=4^2 or 16=16

[@sanjaykamath90210]

0 2,4

[@carmicha]

But there are many other solutions, e.g., if x=y then x^y = y^x as well as 2^4=4^2 .

[@arturradwanski5397]

The second one is actually coverd in a video, when you substitute k=2 you get x=2 & y=4

[@minhhainguyen2671]

❤❤❤❤❤.

[@davidchen275]

x=y

[@homeauburnRaja-hm3gq]

Take log and it leads to y=x

[@tomtke7351]

x^y = y^x clearly x, y = 1 is one solution as is x, y = 0. just by inspection in fact... any x = y solution works hmmm this author explores y = kx such that the ratio k =y/x is different than 1. x^k = xk k = x^(k-1) k^(1/(k-1))=x but y=kx so y=k[k^(1/(k-1))] so@ 7:03 in video something I don't understand happened as the author comes up with y=k^(k/(k-1)) is wrong? and I believe y=k^(2/(k-1)) is right? y log(x) = x log(y)? y/x = log(y)/log(x) or?? y÷(log(y)) = x÷(log(x)) refresher log.10(a) = b means 10^b=a

[@kalyanbanerjee3490]

😄😄😄

[@Handle-nj5fs]

Time waster!!! 1:18 Couldn't you cancel the power of x from both the sides and make it x^k=kx earlier? Reached x^k=kx at 4:03. Square roots was not at all needed for that.

[@davidz1436]

Vary complicate for a problem with a lot of solution, X = Y.

[@santoshkumarpradhan5213]

X=Y this is the Right answer. Itina badha Chadha ke kya jarurat he

[@sigillumdei887]

Obviously when I have seen the thumbnail in video I realized that x=y=0. No more complications than that any number rised to power 0 is 1. This was 1 second solve then I realized there are infinite number of solutions. If this is a olimpiad question then my medium math knowledge seems advanced :))) no way can be a olimpiad one.

[@ronaldnoll3247]

Another result is x=1 and y=1

[@user-ig7nj1xb5r]

i don't understand. why y=kx ( k= constant) ?

[@Mamtamaam]

Yes

[@AlainPaulikevitch]

very good question this is the most important step in the resolution and it is not even explained. it actually allows finding one family of solutions but the logic behind arbitrarily choosing it is not explained. why not try something like cos(x) = k / y (to give a silly arbitrary alternative). My point being that when solving such problems one has to justify their solution as being complete, and in this video this is not done, possibly because the trivial solution anybody can guess, however as far as i can figure there are 3 families of solutions. x = y for x and y >= 0 is a solution y = k pow(k/(k-1)) and x=k pow(1/(k-1)) for k > 0 and k not equal to 1 is another solution in the real numbers set And there is a third potential solution in the realm of complex numbers when either (or both/not sure) x or y are such that ln(x) = x or ln(y) = y (i have no idea how this is solved but googling it leads to people solving it in the complex number set) I somehow feel that the expected solution to this problem is to find an equation for the second solution which is not a parameterized equation but rather in the form y = f(x) which would be nicer, but have no reason to believe that this is possible, neither can i surmise that it isn't as not having found it does not mean it isn't there. This wouldn't be an olympiad question if the solution didn't allow for further refinements to differentiate the candidates.

[@mmagic5753]

what if x = 1, and y = 1? r u take serious as a math tutor on youtube? dont missleading ur watcher.

[@sudhirsharma7966]

Why not 5 not 7 hahahahahah

[@mithrasrevisited4873]

X = 1 and Y = 1. Am I missing something?

[@GeoRedtick]

Valid for any x=y.

[@igortchernowitzer927]

2/4=4/2

[@GuiAguirrezabal]

Absurdo. Tanto trabalho ... Qq x=y; y sendo potencia de x ...

[@xyzw2468]

Otra solucion es: y = x ProductLog[ - (Log[x]/x) ] )/Log[x]

[@valeryshebaldenkov9326]

Сорри - это чисто русское решение... дебильное...

[@user-zj9vt9el3w]

X=4 ; Y=2

[@kailasshirore22]

Edpat

[@renebrienne1862]

Sorry, , but stupid ans unuseful trial of solution !!!!! Thé simple solution ( in fact an infinity)IS, of course x= y.

[@rajroy44]

(silly) and even this silly example is wrong.

[@SuperAnangs]

x= 0, y = 1

[@uduehdjztyfjrdjciv2160]

Only zero to power of zero can be one, other powers of zero are zero. One only in power of infinity can be e. Other powers of one is one.

[@keescanalfp5143]

in your case, are we really ready to admit that 1⁰ = 0¹ so 1 = 0 . ? yeah it's almost perfect . we would not dare to write down this . but yeah you really earn your teacher's heart . good luck .

[@niranjanchakraborty1139]

Ans=x=y.

[@northpole1674]

You don't have any idea of mathematics.