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
@nyonkavincenttafeli7002
Жыл бұрын
Perfect. Absolutely correct
@HenriqueSantos-xd1eg
Жыл бұрын
Why y=Kx?
@wangpercy2765
Жыл бұрын
@@HenriqueSantos-xd1egjust because😂
@raderadumilo7899
Жыл бұрын
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.
@falahalfadhel185
Жыл бұрын
y=kx , k = constant , that mean y Directly proportional to x therefor no problem
@irahartoch1075
Жыл бұрын
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....
@reznikvolodymyr8145
8 ай бұрын
So the problem is not solved by she?
@kereta_Api_CC206
7 ай бұрын
Hahaha
@kereta_Api_CC206
7 ай бұрын
The answer is a graphics y=x
@nicolasottocornola3166
6 ай бұрын
X=2, y=4 is her solution for k=2.
@ExploringLife333
5 ай бұрын
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.
@klauscosmin
Жыл бұрын
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)
@EnoughIsEnough571
Жыл бұрын
This is the problem with byhearting the math solution rather than using logic.
@alexyuri_94
Жыл бұрын
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
@anandakundu9317
Жыл бұрын
I was wondering the same think bro
@Alan-sv6ym
Жыл бұрын
@@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
@valeriotamellini2006
Жыл бұрын
You are right. SQRT(2) is useless.
@crmn_tv
Жыл бұрын
x = 1, y = 1 can also be a solution (if there is no other constraints such as x != y)
@dnagpal
11 ай бұрын
That was my first thought. x=1, y = 1, then x = 0 and y = 0 and basically x = y
@stuarts4770
Жыл бұрын
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.
@falahalfadhel185
Жыл бұрын
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
@aberro72
Жыл бұрын
When k=1 => the solution is x=y=e. That assuming y=k * x. In general, when x >0 and x
@weeblyploonbottom810
11 ай бұрын
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.
@aberro72
11 ай бұрын
@@weeblyploonbottom810 Silly "things" on internet for sure.... 😁😁
@DaHaiZhu
Жыл бұрын
Seems there are infinite solutions where x = y, since that condition was not excluded.
@enki354
Жыл бұрын
That's what I say
@emaildomagno
Жыл бұрын
x=y
@orchestra2603
Жыл бұрын
@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.
@hlindstrom
Жыл бұрын
Yes, that seemed obvious, x=y. Why not just answer that due to symmetry x = y?
@dariolazzari2415
Жыл бұрын
@@hlindstrombecause there are also infinite solutions where x≠y
@Godeau03
Жыл бұрын
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.
@nikkverma5523
Жыл бұрын
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).
@josephpatti2835
Жыл бұрын
@@nikkverma5523This is what I thought no need to do anything
@Alfanoustv
Жыл бұрын
@@nikkverma5523 (x,x) in general is a solution.
@weeblyploonbottom810
11 ай бұрын
Dude, it was a prank. See my comment above about dihydrogen monoxide
@timm7142
Жыл бұрын
As long as x=y, there are infinite number of roots. 👍👍👍
@chrismcgowan3938
Жыл бұрын
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 :-)
@casperkruger348
Жыл бұрын
The vital part of the question should be y != x. Otherwise, the simple answer is y = x = 1
@earnandlivebetter
Жыл бұрын
x=2, y=4@@chrismcgowan3938
@casperkruger348
Жыл бұрын
@ajaysamanta9661 😶🌫️
@igoranisimov6549
11 ай бұрын
And ironically she did not mention that k may not be equal 1 in their solution
@mariorodriguezruiz8519
Жыл бұрын
Why assuming y=kx? That only gives a family of solutions
@abdullahmoh1732
Жыл бұрын
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.
@anotherelvis
8 ай бұрын
If x!=0, then you can always define k=x/y. And then you can look for solutions corresponding to each value of k.
@bobbob-gg4eo
8 ай бұрын
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?
@anismanjhi4342
Жыл бұрын
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.
@Mr_Basketball95
11 ай бұрын
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.
@luciusluca
11 ай бұрын
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.
@MsRa3d
Жыл бұрын
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
@dagoonsg9634
Жыл бұрын
I thought u where coming good until ur 1st new equation, that right side doesn't add up I think
@thasicommunitiyheatlhcarec3459
Жыл бұрын
Dad carries his son on his shoulder and when the dad becomes older, the son carries his dad on his shoulder: PROVED
@jevilsugoma1743
Жыл бұрын
Lmao
@theupson
11 ай бұрын
f(u) = logu/u on u>1 has two continuous monotone branches {(1
@theupson
11 ай бұрын
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.
@aapiElder
Жыл бұрын
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.
@bernardseffah2886
Жыл бұрын
Logically, if x=y, then any number satisfies the equation
@123prova
9 ай бұрын
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
@SomeoneCommenting
Жыл бұрын
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".
@Thirukkural-Stories
Жыл бұрын
You have not found the values of k. You are substituting an arbitrary vaue of 3 to k. This is not a mathematical solution.
@josesszwec835
Жыл бұрын
What about x=y=1, x=y=2 and x=2,y=4?
@dbdb7745
Жыл бұрын
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.
@sniperwolf50
7 ай бұрын
Except 0, as 0^0 is undefined
@tube102000
Жыл бұрын
Apologize for my poor math skills, but doesn’t this have infinite solutions (x=y)?
@weeblyploonbottom810
11 ай бұрын
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
@redroach401
9 ай бұрын
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.
@musicsubicandcebu1774
Жыл бұрын
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)
@longcours
Жыл бұрын
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 ?
@jevilsugoma1743
Жыл бұрын
Same question
@OptimusPrime-vg2ti
11 ай бұрын
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.
@berxillos
Жыл бұрын
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
@bhaskarps
Жыл бұрын
Why to take square root? One get same result without taking square root.
@Rick_MacKenzie
Жыл бұрын
That was absolutely bizarre. There is nothing in the original equation to even suggest taking square root. There was no two in the exponents.
@giorgioevangelisti1369
Жыл бұрын
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.
@pcrtwentekanaal8458
Жыл бұрын
Exactly my thought! Taking the square root makes no sense.
@jeancharleskorta7633
Жыл бұрын
A cube root or any other power other than zero would make it look even more "smart" albeit completely unnecessary!
@dmitryr5453
Жыл бұрын
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...
@hermenkamya729
Жыл бұрын
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.
@orchestra2603
Жыл бұрын
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
@arielsinardi2626
Жыл бұрын
Qué pasaba si en vez de elegir K=3 se elegía otro valor?
@marcfirst9341
Жыл бұрын
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...
@cheikh7036
Жыл бұрын
Along with the case where x=0, before using the 2/x simplification even though it's obvious
@falahalfadhel185
Жыл бұрын
when saying y=kx and k=constant that mean k>0 ,I think the question is closer to physics than mathematics
@PugganBacklund
Жыл бұрын
the first square-root seamed unnessesery,
@pcrtwentekanaal8458
Жыл бұрын
Indeed
@DavidLealvalmana
Жыл бұрын
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.
@albajasadur2694
11 ай бұрын
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.
@loong111
Жыл бұрын
Its a single equation with 2 variables. There are an infinite number of x,y values
@michaelhartmann1285
10 ай бұрын
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.
@antonyqueen6512
Жыл бұрын
What was the step of making the square root for? Just complicating things or just making the video longer!??😂😂😂
@meghraj1234567890
Жыл бұрын
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
@ardeshirhaidarbaigi5336
Жыл бұрын
Hello and Thanks. I do not understand how do you choose "...If k = 3, ..." at 5:18 . Can you please explain it
@danielderoudilhes4413
Жыл бұрын
No meaning.
@JeffreyBue_imtxsmoke
9 ай бұрын
Isn't this equation satisfied for all X=Y? Maybe excluding zero? Is 0 to the power 0 even defined?
@JasonTse
Жыл бұрын
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.
@kameshvengatta4381
Жыл бұрын
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
@JadeDragon407
Жыл бұрын
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!
@weeblyploonbottom810
11 ай бұрын
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.
@JadeDragon407
10 ай бұрын
@@weeblyploonbottom810 I feel like the matrix just had an uh oh. >>:=p
@Anders01
Жыл бұрын
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.
@SpacePhys
Жыл бұрын
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.
@OptimusPrime-vg2ti
11 ай бұрын
@@SpacePhys At 3:30 you can't divide both sides by k-1, without implicitly assuming k ≠ 1.
@howardfagan9177
Жыл бұрын
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.
@darkbluemars
8 ай бұрын
I find your videos fascinating. I don't know why I keep watching it before I sleep though.
@matsonnerby
Жыл бұрын
This should be true for every value where x=y
@cheikh7036
Жыл бұрын
And this is in total contradiction with the proposed solution where k=1 is excluded
@ciba0318
Жыл бұрын
Yes, x=y is a solution only that x and y not equal to zero
@robertsalazar2770
Жыл бұрын
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?
@dearkarnataka
Жыл бұрын
I never understood math when you simply added some letters and then eliminated them later
@hitest8925
6 ай бұрын
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.
@jllaury75
Жыл бұрын
Al ojo: (2^4)=(4 ^2) X =2 ; y =4 ó vicecersa.
@michaelhuppertz6738
4 ай бұрын
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
@memineown4415
Жыл бұрын
Interesting...so instead of saying "there are many paths to solve this" she just forges ahead with no explainations? Why did she choose this path? Then, reminds us of a law but doesnt explain why she needs to add a constant into the equation? Then ends up saying its a proof...so what? We can see its an equation. The value of x is what? And the values of y are...? Thats the question to be answered. K is an irrelavance. Try graphing it.
@EhsanZia-Academi
6 ай бұрын
Thank you Mom for your great videos. Could you please give me the link for these questions references?
@manikbanik4796
Жыл бұрын
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.
@corneliusagu2903
Жыл бұрын
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..
@biodreg1332
Жыл бұрын
Any two real numbers are linearly dependent. dim R = 1.
@fauzan9178
Жыл бұрын
Why take long long calculate, There are variables, we can try different approaches to find possible solutions. Let's start with a few simple examples: If x = 2 and y = 4: 2^4 = 4^2 16 = 16 If x = 4 and y = 2: 4^2 = 2^4 16 = 16 This example satisfies the equation. From these examples, we can see that when x = 2 and y = 4 or when x = 4 and y = 2, the equation x^y = y^x holds true. So, the solutions to the equation x^y = y^x are x = 2 and y = 4 (or vice versa) and x = 4 and y = 2 (or vice versa).
@sajjadali1902
Жыл бұрын
Excellent
@ahlee2
Жыл бұрын
Great 👍
@RookieGamerz-3110
6 ай бұрын
It was going fine until u took k = 3 so randomly.
@jameshenry3530
11 ай бұрын
the solution can also be x=1 and y=1. X to the power 1 = 1; Y to the power 1=1.
@SenChai
Жыл бұрын
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.
@panKiev
Жыл бұрын
Who says that this should be a linear relationship between y and x?
@guruvenon
Жыл бұрын
yeah, i agree with you..
@subham1225
Жыл бұрын
X = 2 , Y = 4
@iceman9678
Жыл бұрын
I thought the clever answer was x = y. It doesn't specify that x can not equal y.
@evilcott
11 ай бұрын
Самый простой ответ x=1, y=1. Ответ чуть посложнее - x=2, y=2. И архисложный ответ - x=3, y=3. И так можно продолжать до бесконечности.
@Massive1986Cava
11 ай бұрын
Can i simply say 2 and 4? With no calculation but the equation is correct
@pedrofajardo8137
Жыл бұрын
You transfomed the original equation
@whosit112
5 күн бұрын
There is absolutely no reason for the square root step. It is entirely superfluous.
@BlakeyHyde
Жыл бұрын
There is a flaw in the reasoning. Why was square roots taken in aroundthe 3rd step? No good reason given.
@ARTEBRAAssocCultural
Жыл бұрын
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...
@pedrocas290791
Жыл бұрын
5) if x is even, the step of power to 2/x result in a modular equation.
@mateuscarvalho846
Жыл бұрын
Very Beautiful Question !!!!!! 👌🏾👍🏾🤙🏾 👏👏👏👏👏
@evgtro8727
Жыл бұрын
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).
@korruskorrowaty5858
11 ай бұрын
The simplest solution is where x=y for that condition any given number is correct.
@ЕвгенийОдинцов-г1т
Жыл бұрын
Well, and what about y is not = kx?
@ermajisetiawan4019
Жыл бұрын
👍👍amazing....
@manojkantsamal4945
8 ай бұрын
Madam 🙏,, you made this problem easy by using an inovative idea..
@maveric3218
Жыл бұрын
HI :-) In 4:34 min, Why does y=K^1+1/(K-1) become y=K^K-1+1/(K-1). I didn`t study math, so I can`t understand this step. If you could explain it, I`ll be thankfull. :-)
@Bubbledragon1
9 ай бұрын
because you are doing the addition, the 1 can be written as k-1 over k-1, as k-1/k-1 = 1, hope that helps \o/
@barakathaider6333
2 ай бұрын
So.... What is the point in the beginning you took square root both sidees and after you raise to 2?
@snmklc160
8 ай бұрын
This is right for every x=y case apart from 0
@TwoMarlboro
Жыл бұрын
but you missed the solution line x=y.
@mostafarageh1647
Жыл бұрын
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
@David-nb4ph
Жыл бұрын
También fácilmente le puedes asignar el valor de 1 a ambas variables y listo.
@davidren2084
Жыл бұрын
it is easy to found the y=x is one of group value,so x and y is able to every no.
@JucLansegers
Жыл бұрын
x=1 y=1 or in short x=1=y no calculation needed. Valid solution?
@rajaijazh
Жыл бұрын
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.
@alexandresalgado8247
Жыл бұрын
this question goes against the basic principle that the number of equations has to be equal to the number of unknown variables
@familytied6976
Жыл бұрын
I don't like the way this solution was executed. It should be considered as a crime.
@dec1aim_flow432
Жыл бұрын
I don’t understand why you impose Y=kx. Was it in the question? I believe you also could have tried y=ax+b or anything. Why a polynomial relation? I don’t get it. It seems very limited as we might lose a lot of solutions
@abdelhakimmoussi6910
7 ай бұрын
Don’t you think that we should mention that K has to be # from 1? Otherwise we can’t put (K-1) in the denominator.
@shivendradwivedi2838
5 ай бұрын
Nice explanation
@LKLogic
5 ай бұрын
Thanks for liking
@shivendradwivedi2838
5 ай бұрын
@@LKLogic welcome
@kimsanov
Жыл бұрын
But k cannot be 0. So all solutions where x = y are missed, though y = kx is valid when k=1
@jst_vjkisoensingh3236
Жыл бұрын
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 😢
@TheSiriusEnigma
Жыл бұрын
The question does not explicitly exclude that x != y. It should.
@AFFB
Жыл бұрын
As there isn't any restriction in command, then the answer is x=1 and y=1.
@nagarajahshiremagalore226
5 ай бұрын
I think this could've been solved without performing the square root in the start.
@tsvetelinpavlov2786
Жыл бұрын
I am not a mathematician, but "lets assume k=3" and assume anything for all the other numbers seems dumb to me.
@tristan583
2 ай бұрын
Yep , the proper thought process lack here , you just assume things
@zedside8106
Жыл бұрын
Why aren’t all ordered pairs x=y from negative infinity to positive infinity solutions?
@billyoung8118
11 ай бұрын
You also have the trivial case where x=y (as long as x, y not equal to zero)
@kumardigvijaymishra5945
Жыл бұрын
5:56 y=kx=3√3 avoids the long way to get to (√3)^3 at 6:35. That saves about 1 minute of every viewer.
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