If you liked this video, then you would probably like this one too. Find all solutions to x^2=2^x (ft Lambert W function) 👉 ruclips.net/video/ndA0sF_0Rwk/видео.html
It's interesting, that if You draw a graph showing x^y=y^x and allow x=y, You get some curve and a ray y=x and the ray intersects the curve in point (e,e). :) Awesome!
This yields the sequence of ordered pairs {(2, 4), (9/4, 27/8), (64/27, 256/81), (625/256, 3125/1024), (7776/3125, 46656/15625), ...}, which lists every rational solution. A simple way to generate it is: x = (1+1/D)^D y = (1+1/D)^(D+1). Thus evidently the rational solutions, when ordered in this way (or equivalently, by the size of the denominator when expressed in least terms) approach the irrational solution x = y = e, where the two parts of the curve x^y=y^x intersect.
Very nice! This certainly gives rational solutions, since 1/(t-1)=D is an integer. Probably it gives all rational solutions, but this requires proving.
The most fun part is when you plug in all the complex numbers and see that they all work. I plugged in t = i and not only did it work, but it also generated a real number. Fabulous.
Actually y^x is always equal to x^y if you can write y^x or x^y as the (b-1)th-root of b ^ b * (b-1)th-root of b. If we say b=3, we have 2nd-root of 3 ^ 3* 2nd-root of 3, which is equal to 2nd-root of 27. And if b=4, we get that the 3rd-root of 4 ^ 4*3rd-root of 4 = 4*3rd-root of 4 ^ 3rd-root of 4
@@daniledenial you just wrote the same thing that in the video. And the goal wasn't to proof x^y = y^x if x = sqrt(3) and y = sqrt(27), but to solve the equation for real x and y with the condition that x is not equal to y.
Here is what I did. First, I took the log of both sides. Thus, yln(x) = xln(y). Then I moved both sides over such that ln(y)/ln(x) = y/x. The problem now looks very simple, its simply saying that both side should be of the same ratio. I then assumed some parameter "c" between both sides of the equality. (We possibly could have done this at the beginning, but that way would have been very difficult). So now we have ln(y)/ln(x) = c and y/x = c. Then we use the linearity of these equations to make our lives very easy. ln(y) = cln(x) and y = cx. Now we simplify the first equation, y = x^c. These two equations imply that x^c = cx. Ah, now both sides have the same base, we can solve for c. x^c * x^-1 = c simplifies to x^(c-1) = c. Taking the (c-1) root of both sides simplifies the equations to x = c^(1/(c-1)). And there you have it! Plug a value into c, and it will yield the x value you need, then multiply that x value by c to get the corresponding y value. This forms the solution space of y^x = x^y
I just let y be dependent on x, so that y = x^p. We get x^(x^p)=(x^p)^x. x^(x^p)=x^(px). x^p=px x^(p-1)=p x=p^(1/(p-1)) Plugging in any value for p we get solutions
He knows that. He is looking for values where exponentiation is commutative, that's why he assumes that x^y = y^x (associativity is something else btw)
This was very high on the "cool thermometer"! In college in the late 1960's, a few of us math majors were investigating solutions to that equation, just out of curiosity. While we concluded that (2,4) and its symmetric partner (4,2) give the only integer solution where x≠y, we never hit on this parametric formula. We did find that if you graph the solution set in the first quadrant, it consists of the line y=x (obviously), along with a curve that loosely resembles the rectangular hyperbola xy = 1, but shifted by (+1,+1); and that the two lines intersect at (e,e) [which is also pretty cool]. Of course, the graph is symmetric about y=x; and as either x or y → ∞, the other → 1⁺. So this means that the curve mentioned above, more closely resembles (x-1)(y-1) = (e-1)² ≈ 3 Fred
You're welcome! I'll be on the lookout for that. It *is* kind of an intriguing topic. And having put it aside so many years ago, I was glad to see your parametric solution which, for one thing, makes it a whole lot easier to plot... Fred
To me, a highschool student who has only done a watered-down version of Calc 1 in a course, because my country doesn't care about Math, I saw this problem as impossible from the title. But as you started with the second solution, I realized how elegant and connected Math can be sometimes. It just shows how well you can actually communicate your thoughts and explain them well. I'm more of a physics guy, but a deep connection to Math is quite important to me, not just getting something right. You've helped me through that. Thank you for your efforts, and hopefully, with enough practice as it is, I'll be able to view seemingly impossible problems the same way you do! Totally possible!
Nah nah, I agree, absolutely zero countries can function without maths, maths is a quintessential subject in every country for basic functionality over stock exchanges to expeditions, no country doesn't value maths, either the original post is lying and he is just bad and not passionate about maths, or he lives in the tribelands of somalia
If you introduce y = x^t instead of y = tx @13:58 you will get to the same place just as fast but maybe just a bit easier. Good videos, very enjoyable.
Actually, one of the solutions you find in your first method is the value y itself (in your example x = 3 is a solution), which you excluded. On way to study the number of solutions of this equation is to rewrite it by taking logarithms (you get y ln(x) = x ln(y) ) and rearranging to a more symmetrical : ln(x)/x = ln(y)/y. So you have to study when the function f(t)=ln(t)/t takes the same value twice. Since f is increasing on ]0, e] from -∞ to 1/e and decreasing on [e,+∞[ from 1/e to 0, it can be shown that for any x in ]1,e[, there is exactly one y in ]e,+∞[ such that f(x) = f(y).
I actually thought about this problem once, and I was considering natural number solutions for x and y, and I got stuck. After you showed the function x=t^(1/t-1), I graphed it and found that it asymptotes to x=1 as t->+infinity [(+infinity)^0=1 perhaps?], and to t=0 as t->0+. The graph only crosses one lattice point at (2, 2). x is already smaller than 2 for t>2, so no more whole number solutions can exist for t>2. And t=1 is undefined, the limit approaches e=2.718... as t->1 t^(1/t-1) = n->0 (1+n)^1/n = k->+infinity (1+1/k)^k. Therefor, 2^4=4^2 is the only natural number solutions for x^y=y^x.
For x=t^(1/t-1), t=2 gives the only whole number solution of x=2^[1/(2-1)]=2^1=2. For y=t^(t/t-1), t=2 gives the only whole number solution of y=2^[2/(2-1)]=2^2=4. 2^4=4^2 is the only equation with whole numbers in the form of x^y=y^x, x≠y.
9:26 At this point, you divide by t before plugging in x, but then you multiply by t before solving for y. Can't you just plug in x directly and skip a couple steps here?
Great video! This is one of my favorite equations of all time. Try to prove that 2^4 = 4^2 is the only Integer solution in the next video. It's a fun proof
Please write it, it seems interesting - one of those thing's u have a hunch of but just find difficult to prove. He accidentally made a mistake in saying 3 & 27 work. Thx in advance
Actually no. It might seem like it, but actually see it to completion. Example case (3,3³=27): 3²⁷(3 multiplied 27 times) is NOT equal to 27³(three 3s multiplied 3 times, hence 3 multiplied 9 times) It all comes down to the solutions of that xᵗ = tx, which I believe to have only (x=2, t=2) as the only integer solutions, related to 2² = 2(2). 35cut may know how to prove it
Hi. Thanks for your math videos! I have a question: around 7:25 you make a simplification which basically is: if a^b = c^b, then a = c. But if b is even, a = -c is also a solution, and this simplification may make you lose solutions. So the proof would have to be completed with the possibility that a = -c and b = 2k, with k an integer, right?
11:53 using a number “b” for t OR the reciprocal of that same number (1/b) will generate the same x and y values. take t=2 for example it will generate x=2 and y=4, now use the reciprocal t=1/2 and you will have x=4 and y=2 y=x^t and x=y^-t
Unlike Europe, Asia and Americas generally use "." For decimal points, like 3"."27 and rarely use "," (except for grouping large numbers like 2,000,000) so yeah it's a bit different
its just that x=3 and y=27 dosnt work at all or i dont get it... 19683 =/= 7625597484987 witch whould be the results for x to the power of y and y to the power of x in this case
@@gepard1983 I was confused too, but I figured it out. Those points are the two solutions where x^3=3^x. The 3,27 point is the trivial one, where 3^3 = 3^3=27. The other one is where 2.4781^3 = 3^2.4781 = 15.2171.
i have a question sir...in this case we assumed Y=tX, so there can be more linear equation to be assumed, and that's how we can get different expressions for the variables.don't we?
At 9:50 when you're solving for y in terms of t, since you discovered the expression for x, can't you just plug in that expression into the equation that's solved for y without any manipulation before hand? ie., if x = t^(1/(t-1)) and y = tx, then doesn't it just follow through straight plug in that y = t ( t^(1/(t-1)) ) ? Small optimization.
what about x^n+y^n = x^y for what x,y,n is it true for positive integers? what about negative ones? btw: can you use a head microphone so you do not have to hold 2 pencils in one of your hands?
You could save a step or two in the third column by starting at y= xt. Also this parametric form shows that the only integer solution occurs at x=t=2 , y=4. Also note that replacing t with 1/t swaps x and y. So all solutions can be found choosing t>1 ie y>x . Further insights may be obtained by considering the graph of ln(t) against lnx and lny
I thought about it, when t < 0, it does not work, x^y will be negative and y^x will be positive, but they have the same decimals. You just add |these| and its okay tho, its a piece of art!
What’s cool is if you plot x^y=y^x you get what looks like two graphs super imposed on top of each other. You get the line y=x for obvious reasons, but you also get this asymptote like bit. But from that you can show that 2 and 4 are the only integer solutions (I’m only considering real numbers, not sure if you could have some complex number solutions with integer real and imaginary parts)
Assuming that x,y are positive we can write lnx/x = lny/y or f(x)=f(y) (1) , where f(x)=lnx/x. Through the graph we can see that (1) has infinite solutions (x,y) if x belongs to (1,e) and y belongs to (e, inf.) and vice versa.
I don't know if you have done a video on this, but you should do a proof that sinh(x) and sinh inverse only intersect at the origin (without graphing tools). It was a problem from my calculus 3 class and I was the only one who got it :P
Essentials Of Math I think I will consider f(x)=sinh(x)-arsinh(x) then use IVT for showing a root then use derivative for showing it's always increasing.
Using our beloved W function, we can solve for x with given y>0: x=e^(-W(-ln(y)/y)) Note that -ln(y)/y is between -1/e and 0, which means the W "function" has actually two different values, one of which gives us the boring x=y result.
At 7:22 when you raise b/s by 1/x , you get x^t = tx which is not always true ...for example, (-2)^2 = 2^2 does not mean that -2 = 2......Please explain.
There is also an explicit formula whit which you can find the y for a given x: If y^x = x^y, then y = -x*W(ln(x)/x)/ln(x). This can easily be verified by trying to transform the right expression into the left one.
btw if you do this with natural logs: x^t-1 = t take ln of both sides ln(x^t-1) = ln(t) put power to the front t-1ln(x) = ln(t) divide both sides by t-1 ln(x) = ln(t)/t-1 put e as a base to both sides e^(ln(x)) = e^(ln(t)/t-1) cancel out e and ln x = 1/t-1
You can also write x as a^b and, then you have x^y = (a^b)^y = a^(by) and it's your another form of this, e.g. sqrt(27)^sqrt(3) = (sqrt(3)^3)^sqrt(3) = sqrt(3)^3sqrt(3) = sqrt(3)^sqrt(27).
I did it in a slight different way. The difference shows up to 8:49. Then i use the ln function: (t-1)lnx=lnt lnx=(lnt)/(t-1) x=e^[(lnt)/(t-1)] But i got a different result. Where did I go wrong? Or that is simply a different way to write that?
Great solution! (Should I do this? Eh, it’s a rare opportunity. No offense though, you’re still WAY smarter than me. See below for his mistake) At 9:30 just multiply tx through; that’s exactly what you did after wasting 30 seconds and adding unnecessary confusion.
I have a story with this problem. I had a crush on a girl I've met days ago and I discovered that she had a boyfriend which got me very sad. To distract myself I started thinking about the powers of two and noticed that 2^4 was equal to 4^2 and I was shocked. Decided to try to solve it. Spent two months trying to solve it. Completely forgot about the girl. Since then I have a new crush : solving math problems. Thank you for being an inspiration to me and helping me pursuit my dreams related to math. I'm in high school but I already know calculus because I loved watching your series "math for fun". Love from Brazil
About 4:10 wouldn’t there only be one solution because X can’t equal Y? And in the context of the equation you’re saying that when X = 3, and X^3 = 3^X, when you plug in X it would be 3^3 = 3^3 . I could be wrong tho but I just wanted to point that out…
7:26 This is a big mistake. Not only you assumed x isn't 0; but there's also many exceptions , such like (x^2)^(1/2)=((-x)^2)^(1/2). By your simplification, x=-x. See where you are wrong?
If you take any number for y, you can always take x=y^y! because you will have y^y^y on bouth sides... like 3 and 27 as shown in the graf-solution, you can do it with 5 and 3125 or any number
Carefully picked rational values lead to nice solutions. Say, anything of the form t = (a + 1) / a. Take a = 3. Then we have x = t^(1/(t - 1)) = ((a + 1) / a)^(1/(1/a)) = (a + 1)^a / a^a, and similarly y = (a + 1)^(a + 1) / a^(a + 1). So they look kind of messy, but plugging in numbers makes them much simpler. (2, 4), (9/4, 27/8), (64/27, 256/81), and so on. You could play around with other parametrized rational solutions that happen to work well and see what you get.
7:58 Notice that when he divides by X, he asumes that X can't be ZERO, which makes a lot of sense since X is Real and if X= 0, the equation would be: 0^y = y ^0 and we all know that 0 to the power of any number (except zero) is 0 and every number to the power of 0 is 1, so we would get an absurd expression: 0 = 1 if X or Y is equal to zero
Philipp Huber while you can still solve it using the lambert w function x^2=e^x x^2 e^(-x)=1 xe^(-1/2x)=±1 -1/2xe^(-1/2x)=±1/2 -1/2x=W(±1/2） x=-2W(±1/2) while W(-1/2) is complex so the real solution of x is-2W(1/2)
If you set x1. Setting t>1, you can prove there are no integer solutions by showing t^(1/(t-1)) only has an integer solution for t=2, which is EZPZ. Pretty cool : D
Your second solution is nice for proving that 1^infinity is indeterminant. As t gets very large (approaches infinity), 1/(t-1) will approach 0 => x will approach 1, and t/(t-1) will will approach 1, which means y will approach infinity. You'll get a conclusion that 1^infinity = infinity, which is an alternative to the intuitive solution that 1^a = 1 for all a.
Finding complex solutions is boring ;) Try solving it in whole numbers ;) Another fun thing to do would be to prove the identity you found (that `√3^√27 = √27^√3`) algebraically instead of depending on Wolfram ;> (because Wolfram can fool you, and it happened before, remember? :q )
I don't understand the proof. You start with the assumption that y is a multiple of x ie y=tx. From there you generate real pairs as functions of t. But what if your initial assumption is not correct? What if you start with an alternate assumption: for example that y is an exponential power of x?
Nick Kravitz well in that example with the exponentiation, that is covered by that case, since y=x^t would mean y=x^t-1 * x, and x^t-1 is just some constant (I will write it c) so y = c x which is the assumption. The assumption is fine as long as we assume that both x and y are not 0. Since t=y/x will always satisfy y=tx.
When we say "multiple", we usually mean an integer multiple, but in this case we never specify that t has to be an integer. So in fact, we can always write y=tx, unless x is 0.
Nick Kravitz - he is not saying that y = tx for a _constant_ value of t. He is actually parametrizing y = f(t) and x = g(t) in such a way that f(t) = g(t)t.
If you liked this video, then you would probably like this one too.
Find all solutions to x^2=2^x (ft Lambert W function) 👉 ruclips.net/video/ndA0sF_0Rwk/видео.html
X € R /X#0
Your ability to effortlessly switch between markers is majestic.
Thank you!!!
Nice
Manny Paul A skill you can’t help but learn when you teach lol
@@DarthAlphaTheGreat
Xxx
@@amirnuriev9092 was
It's interesting, that if You draw a graph showing x^y=y^x and allow x=y, You get some curve and a ray y=x and the ray intersects the curve in point (e,e). :) Awesome!
Yay!!Yay!!
thus it shows every possible x and y value where x^y = y^x
I think the more interesting potion of the graph is the discontinuous section where one of X or Y is negative.
OMG this is why I love the internet
It should intersect e,e and pi pi too
It should intersect every point on x=y
Use this to impress girls. LOL. I’ll let you know if it works for me.
GreenMeansGO : )
LOL
How did it go?
It works with girls!
@@davidbrisbane7206 what age-group ?
If you let t = (D+1)/D, where D is integer, you can find all solutions that do not contain radicals. Example: t = 3/2 gives x = 9/4 and y = 27/8.
This yields the sequence of ordered pairs {(2, 4), (9/4, 27/8), (64/27, 256/81), (625/256, 3125/1024), (7776/3125, 46656/15625), ...}, which lists every rational solution. A simple way to generate it is:
x = (1+1/D)^D
y = (1+1/D)^(D+1).
Thus evidently the rational solutions, when ordered in this way (or equivalently, by the size of the denominator when expressed in least terms) approach the irrational solution x = y = e, where the two parts of the curve x^y=y^x intersect.
Very nice! This certainly gives rational solutions, since 1/(t-1)=D is an integer. Probably it gives all rational solutions, but this requires proving.
I love these no-effort thumbnails, please keep them going 😂
Casey Roberson lol : )
They legendary.
@@15schaa They're*
@@RubyPiec shut
@@wilfriedsteinbach8700 up
"Don't be too crazy.." *maniacal cackle* oh the numbers I'll produce!!
The most fun part is when you plug in all the complex numbers and see that they all work. I plugged in t = i and not only did it work, but it also generated a real number. Fabulous.
"Don't be too crazy"
Me: puts t=1 *cackles maniacally*
Actually y^x is always equal to x^y if you can write y^x or x^y as the (b-1)th-root of b ^ b * (b-1)th-root of b.
If we say b=3, we have 2nd-root of 3 ^ 3* 2nd-root of 3, which is equal to 2nd-root of 27.
And if b=4, we get that the 3rd-root of 4 ^ 4*3rd-root of 4 = 4*3rd-root of 4 ^ 3rd-root of 4
I actually don’t understand how it takes 13 minutes to proof that x^y = y^x if x=squareroot of 3 and y= squareroot of 27
@@daniledenial you just wrote the same thing that in the video. And the goal wasn't to proof x^y = y^x if x = sqrt(3) and y = sqrt(27), but to solve the equation for real x and y with the condition that x is not equal to y.
@@Amoeby Yeah, but there Are More real Solutions
@@daniledenial of course. But that was the example in the video.
Parametric generator for x^y = y^x, wow
wow
Wow
I was like wow when I saw it too!
I took 10 courses in math in college en route to a math apps minor and your videos still amaze me.
Here is what I did. First, I took the log of both sides. Thus, yln(x) = xln(y). Then I moved both sides over such that ln(y)/ln(x) = y/x. The problem now looks very simple, its simply saying that both side should be of the same ratio. I then assumed some parameter "c" between both sides of the equality. (We possibly could have done this at the beginning, but that way would have been very difficult). So now we have ln(y)/ln(x) = c and y/x = c. Then we use the linearity of these equations to make our lives very easy. ln(y) = cln(x) and y = cx. Now we simplify the first equation, y = x^c. These two equations imply that x^c = cx. Ah, now both sides have the same base, we can solve for c. x^c * x^-1 = c simplifies to x^(c-1) = c. Taking the (c-1) root of both sides simplifies the equations to x = c^(1/(c-1)). And there you have it! Plug a value into c, and it will yield the x value you need, then multiply that x value by c to get the corresponding y value. This forms the solution space of y^x = x^y
I just let y be dependent on x, so that y = x^p. We get x^(x^p)=(x^p)^x.
x^(x^p)=x^(px).
x^p=px
x^(p-1)=p
x=p^(1/(p-1))
Plugging in any value for p we get solutions
Adrien Grenier you get the same thing
EDIT: Ohhh, okay I completely misunderstood what you were trying to do.
He knows that. He is looking for values where exponentiation is commutative, that's why he assumes that x^y = y^x (associativity is something else btw)
But how is x^(x^p)=x^(px) equal to x^p=px?
@@nikogruben9573 Both side taking logarithm x-based
Just when I thought there was nothing left for Asians to beat me at, this dude starts writing with two pens in one hand.
never knew this was an asian thing. when i was in school holding a pencil and pen in one hand was normal.
You can't beat us Asians. No one can.
Musicians rule: there is always an asian better, and younger than you. Whether you are asian or not.
This was very high on the "cool thermometer"!
In college in the late 1960's, a few of us math majors were investigating solutions to that equation, just out of curiosity.
While we concluded that (2,4) and its symmetric partner (4,2) give the only integer solution where x≠y, we never hit on this parametric formula.
We did find that if you graph the solution set in the first quadrant, it consists of the line y=x (obviously), along with a curve that loosely resembles the rectangular hyperbola xy = 1, but shifted by (+1,+1); and that the two lines intersect at (e,e) [which is also pretty cool].
Of course, the graph is symmetric about y=x; and as either x or y → ∞, the other → 1⁺.
So this means that the curve mentioned above, more closely resembles
(x-1)(y-1) = (e-1)² ≈ 3
Fred
ffggddss thanks Mr. Fred! Part 2 is coming soon : )
You're welcome! I'll be on the lookout for that. It *is* kind of an intriguing topic.
And having put it aside so many years ago, I was glad to see your parametric solution which, for one thing, makes it a whole lot easier to plot...
Fred
Are powers commutative? Lets just try some values :-P
1, 1 --> 1^1 = 1^1 = 1 --> yes
2, 2 --> 2^2 = 2^2 = 4 --> yes
2, 4 --> 2^4 = 2*2*2*2 = 4^2 = 4*4 = 16 --> yes
2, 3 --> near enough
maybe something a bit less round
sqrt(3), sqrt(27) --> yep
Powers are commutative :-P :) Indisputable proof.
Alan Tennant a simpleton's fallacy.
"indisputable" lol
Alan Tennant love it!!!!
a^(ln(b))=b^(ln(a)) , this way you can commute ...
That's just cherrypicking XD
To me, a highschool student who has only done a watered-down version of Calc 1 in a course, because my country doesn't care about Math, I saw this problem as impossible from the title. But as you started with the second solution, I realized how elegant and connected Math can be sometimes.
It just shows how well you can actually communicate your thoughts and explain them well. I'm more of a physics guy, but a deep connection to Math is quite important to me, not just getting something right. You've helped me through that. Thank you for your efforts, and hopefully, with enough practice as it is, I'll be able to view seemingly impossible problems the same way you do!
Totally possible!
Where do you live?
@@diptoneelde836 he lives his own Dreamland where he thinks his own short comings are his country's mistakes. He needs psychology not maths.
@@pigeonlove you have serious problems
plo Judging by the fact you went out of your way to be a douchebag, I think you need psychology.
Nah nah, I agree, absolutely zero countries can function without maths, maths is a quintessential subject in every country for basic functionality over stock exchanges to expeditions, no country doesn't value maths, either the original post is lying and he is just bad and not passionate about maths, or he lives in the tribelands of somalia
If you introduce y = x^t instead of y = tx @13:58 you will get to the same place just as fast but maybe just a bit easier. Good videos, very enjoyable.
7:55 That’s what my old maths teacher in high school used to say, when we would suggest a harder way to solve a problem.
Actually, one of the solutions you find in your first method is the value y itself (in your example x = 3 is a solution), which you excluded. On way to study the number of solutions of this equation is to rewrite it by taking logarithms (you get y ln(x) = x ln(y) ) and rearranging to a more symmetrical : ln(x)/x = ln(y)/y. So you have to study when the function f(t)=ln(t)/t takes the same value twice. Since f is increasing on ]0, e] from -∞ to 1/e and decreasing on [e,+∞[ from 1/e to 0, it can be shown that for any x in ]1,e[, there is exactly one y in ]e,+∞[ such that f(x) = f(y).
What a great introduction to solving equations with parametrics!
Best way to impress your girls 13:17
Bryan Hart an ad?
I don't get it
Best would be to clear your acne...
looks fucky and here's sexy
The equation e^x=x^e only have the solution x=e. That's the only positive number with this property.
Yes! : )
personsname0 The equation 71^x=x^71 also has a solution x=1.06609898594648539. My equation doesn't have another solution..
thought you were talking bout integer solutions, serves me right for being a smarmy
personsname0 of course I talked about the integer e ;)
lol... oh man, think i might be actually losing my mind
Really interesting. I really like the explanation. Though, when you raise [x^(tx)]^(1/x) x≠0! A big shout out from Argentina!
Bautista Bauto98 thanks!!!!
me: oddly satisfying video
others: nerd
r/iamverysmart
@@fyukfy2366 who else watches this besides nerds though?
@@fyukfy2366 totally what I was thinking
@@tweedyburd007 this is not middle school anymore
@@tweedyburd007 true..
13:04 Sets t = i ; sees into the future.
x^y=y^x is one of my favourite equations
Yay!!
I actually thought about this problem once, and I was considering natural number solutions for x and y, and I got stuck. After you showed the function x=t^(1/t-1), I graphed it and found that it asymptotes to x=1 as t->+infinity [(+infinity)^0=1 perhaps?], and to t=0 as t->0+. The graph only crosses one lattice point at (2, 2). x is already smaller than 2 for t>2, so no more whole number solutions can exist for t>2. And t=1 is undefined, the limit approaches e=2.718... as t->1 t^(1/t-1) = n->0 (1+n)^1/n = k->+infinity (1+1/k)^k. Therefor, 2^4=4^2 is the only natural number solutions for x^y=y^x.
Jack Ren 2 is the only integer solution as t-1 | t can only happen if and only if t = 2
For x=t^(1/t-1), t=2 gives the only whole number solution of x=2^[1/(2-1)]=2^1=2.
For y=t^(t/t-1), t=2 gives the only whole number solution of y=2^[2/(2-1)]=2^2=4.
2^4=4^2 is the only equation with whole numbers in the form of x^y=y^x, x≠y.
The equastion x^y=y^x on his own is pretty easy. The answer is x=y. But with the data he gave it is a realy hard equastion.
9:26 At this point, you divide by t before plugging in x, but then you multiply by t before solving for y. Can't you just plug in x directly and skip a couple steps here?
L1N3R1D3R Ahhhhhhhh loll!!! I didn't see that
Hahahaaa
Haha yep!!
Noticed that as well lmaoo
This guys enthusiasm as he takes us through his story brings unexpected amounts of joy into my life
Great video! This is one of my favorite equations of all time.
Try to prove that 2^4 = 4^2 is the only Integer solution in the next video. It's a fun proof
Even tho it's fun it has absolutely no meaning from what i can tell... no real problem would lead to this equation, i guess?
y = x^x or x = y^y
It will work
Please write it, it seems interesting - one of those thing's u have a hunch of but just find difficult to prove. He accidentally made a mistake in saying 3 & 27 work. Thx in advance
Actually no. It might seem like it, but actually see it to completion. Example case (3,3³=27): 3²⁷(3 multiplied 27 times) is NOT equal to 27³(three 3s multiplied 3 times, hence 3 multiplied 9 times)
It all comes down to the solutions of that xᵗ = tx, which I believe to have only (x=2, t=2) as the only integer solutions, related to 2² = 2(2). 35cut may know how to prove it
Theo Ajuyah oh my bad thx for correcting me
Hi. Thanks for your math videos! I have a question: around 7:25 you make a simplification which basically is: if a^b = c^b, then a = c. But if b is even, a = -c is also a solution, and this simplification may make you lose solutions. So the proof would have to be completed with the possibility that a = -c and b = 2k, with k an integer, right?
0:54 *P O W E R*
In another video u said never trust Wolframalpha
Thumbs up. That y=tx relation was excellent
as always, a very satisfying answer
: )
1:09 the color of the markers on his shirt are the same of the colors of marker he uses.
11:53 using a number “b” for t OR the reciprocal of that same number (1/b) will generate the same x and y values. take t=2 for example it will generate x=2 and y=4, now use the reciprocal t=1/2 and you will have x=4 and y=2
y=x^t and x=y^-t
That impressed me a lot! Seeing such equations solved makes me feel kind of satisfied...
8:58
I like that trick. Idk why I haven't thought about it. I might use it on a logarithim question
Haces la matemática más fácil y entretenida... Más entretenida de lo que de por sí, ya es.
Muchas gracias!
4:09 - i prefer to use ; instead of , between x and y value. For a moment i though that 2nd answer is x= 3.27 and not x=3 y=27
Unlike Europe, Asia and Americas generally use "." For decimal points, like 3"."27 and rarely use "," (except for grouping large numbers like 2,000,000) so yeah it's a bit different
its just that x=3 and y=27 dosnt work at all or i dont get it... 19683 =/= 7625597484987 witch whould be the results for x to the power of y and y to the power of x in this case
@@gepard1983 I was confused too, but I figured it out. Those points are the two solutions where x^3=3^x. The 3,27 point is the trivial one, where 3^3 = 3^3=27. The other one is where 2.4781^3 = 3^2.4781 = 15.2171.
i have a question sir...in this case we assumed Y=tX, so there can be more linear equation to be assumed, and that's how we can get different expressions for the variables.don't we?
At 9:50 when you're solving for y in terms of t, since you discovered the expression for x, can't you just plug in that expression into the equation that's solved for y without any manipulation before hand?
ie., if x = t^(1/(t-1)) and
y = tx, then doesn't it just follow through straight plug in that
y = t ( t^(1/(t-1)) ) ?
Small optimization.
what about x^n+y^n = x^y for what x,y,n is it true for positive integers? what about negative ones? btw: can you use a head microphone so you do not have to hold 2 pencils in one of your hands?
You could save a step or two in the third column by starting at y= xt. Also this parametric form shows that the only integer solution occurs at x=t=2 , y=4. Also note that replacing t with 1/t swaps x and y. So all solutions can be found choosing t>1 ie y>x .
Further insights may be obtained by considering the graph of ln(t) against lnx and lny
How you can divide by x when the x belongs to the |R?
Well, there obviously are no (sane?) solutions for x=0 anyway, so...
I thought about it, when t < 0, it does not work, x^y will be negative and y^x will be positive, but they have the same decimals. You just add |these| and its okay tho, its a piece of art!
I feel so smart that I understood what you did in this video
: )
What’s cool is if you plot x^y=y^x you get what looks like two graphs super imposed on top of each other. You get the line y=x for obvious reasons, but you also get this asymptote like bit. But from that you can show that 2 and 4 are the only integer solutions (I’m only considering real numbers, not sure if you could have some complex number solutions with integer real and imaginary parts)
Really cool vid this makes more excited to learn calculus although most of the questions probably will be mostly the daily grind type unlike this one
4:03 That AaaaHaaa caught me off-guard ngl
I liked this one a lot!
at 10:06, why is y=tx rewritten as x=y/t, since both sides are multiplied by t in the next step
👏 this is so good; so good
#U deserve an applause
Charles David thank you!!!!
Assuming that x,y are positive we can write lnx/x = lny/y or f(x)=f(y) (1) , where f(x)=lnx/x. Through the graph we can see that (1) has infinite solutions (x,y) if x belongs to (1,e) and y belongs to (e, inf.) and vice versa.
I don't know if you have done a video on this, but you should do a proof that sinh(x) and sinh inverse only intersect at the origin (without graphing tools). It was a problem from my calculus 3 class and I was the only one who got it :P
Essentials Of Math no. I don't have it yet. And I think I will need to think hard on that.
blackpenredpen thinking hard is the best! I'll look forward to it!
Essentials Of Math
I think I will consider f(x)=sinh(x)-arsinh(x) then use IVT for showing a root then use derivative for showing it's always increasing.
blackpenredpen you're on the right track, but there is a much simpler function you can use! :)
blackpenredpen
Think about the derivative of the separate functions sinh and arsinh
Everyone needs to learn this formula. Its great
is this considered pre-calculus? how important would parametrics be going into calc 2?
I think it's do-able in precalc. Parametric equations are useful in calc2, 3, and more : )
thanks! and are you at UCB? do you teach there? would be fun to have you
Using our beloved W function, we can solve for x with given y>0:
x=e^(-W(-ln(y)/y))
Note that -ln(y)/y is between -1/e and 0, which means the W "function" has actually two different values, one of which gives us the boring x=y result.
"You can use whatever t you want" mmmh let's take t=1
Did you forget that y ≠ x?
Максим Быков yeah you’re right. If t=1 then y would equal x which isn’t allowed
@@disc_00 check it out again
If t=1 ,then it would be a 1/0 as x's exponent
Think first,comment after
@@tylerchristensen7480 @Максим Быков check it out again
If t=1 ,then it would be a 1/0 as x's exponent
Think first,comment after
@@mohammadfahrurrozy8082 Yes by the limit as t tends to 1 is still defined, this gives x = y = e.
At 7:22 when you raise b/s by 1/x , you get x^t = tx which is not always true ...for example, (-2)^2 = 2^2 does not mean that -2 = 2......Please explain.
There is also an explicit formula whit which you can find the y for a given x:
If y^x = x^y, then y = -x*W(ln(x)/x)/ln(x).
This can easily be verified by trying to transform the right expression into the left one.
PHScience What is W in your equation?
Jonas the lambert W function
I was thinking of the same. But I thought it wouldn't make sense in this context. thank you
Never logarithm when you can find another way.
What's the W function
btw if you do this with natural logs:
x^t-1 = t
take ln of both sides
ln(x^t-1) = ln(t)
put power to the front
t-1ln(x) = ln(t)
divide both sides by t-1
ln(x) = ln(t)/t-1
put e as a base to both sides
e^(ln(x)) = e^(ln(t)/t-1)
cancel out e and ln
x = 1/t-1
setting y = tx to x = y/t was redundant lol
AniPrograms Yeah, That part was totally unneeded. Still a great video.
You have an anime profile picture
@@orcishh setting y = tx to x = y/t was redundant lol
Now its a valid comment thank me later
@@ilprincipe8094 but this kid has an anime profile picture
@@orcishh goddamnit you are right
You can also write x as a^b and, then you have x^y = (a^b)^y = a^(by) and it's your another form of this, e.g. sqrt(27)^sqrt(3) = (sqrt(3)^3)^sqrt(3) = sqrt(3)^3sqrt(3) = sqrt(3)^sqrt(27).
Your accent has improved a lot .
Keep up the good work.
9:23 - i think we don't need to divide both sides by t, substitute x and then multiply by t again; we can just substitute x into y=tx
9:27 You divide by t and then multiply by t one step later ;)
y was already isolated :) Still, cool video!
I did it in a slight different way. The difference shows up to 8:49.
Then i use the ln function:
(t-1)lnx=lnt
lnx=(lnt)/(t-1)
x=e^[(lnt)/(t-1)]
But i got a different result. Where did I go wrong? Or that is simply a different way to write that?
Great solution!
(Should I do this? Eh, it’s a rare opportunity. No offense though, you’re still WAY smarter than me. See below for his mistake)
At 9:30 just multiply tx through; that’s exactly what you did after wasting 30 seconds and adding unnecessary confusion.
To be honest, I am not that smart.
I have a story with this problem. I had a crush on a girl I've met days ago and I discovered that she had a boyfriend which got me very sad. To distract myself I started thinking about the powers of two and noticed that 2^4 was equal to 4^2 and I was shocked. Decided to try to solve it. Spent two months trying to solve it. Completely forgot about the girl. Since then I have a new crush : solving math problems. Thank you for being an inspiration to me and helping me pursuit my dreams related to math. I'm in high school but I already know calculus because I loved watching your series "math for fun". Love from Brazil
I usually watch some videos of yours, but this one seriously made me think "holy shit this is genius" lol
: )))))
About 4:10 wouldn’t there only be one solution because X can’t equal Y? And in the context of the equation you’re saying that when X = 3, and X^3 = 3^X, when you plug in X it would be 3^3 = 3^3 . I could be wrong tho but I just wanted to point that out…
Why not take the natural log of both sides and solve implicitly?
Micah Beiser it wont do anything
Take the natural log of one side over the natural log of x or y and solve from
log(x^y)/log(y)=x
this is my favorite channel. I already loved math but on this channel, I learn things I love.
The 1's cancel out each other. -1+1=0
7:26 This is a big mistake. Not only you assumed x isn't 0; but there's also many exceptions , such like (x^2)^(1/2)=((-x)^2)^(1/2). By your simplification, x=-x. See where you are wrong?
BLACKPENREDPEN #YAY
WE LOVE YOU
Thank you!!!
You can try to find integral solutions to the equation with some different approach
SOOOOOOOOOO COOOOOOOOOOOOOL YEEEEEY! Thanks for being awesome with math
At 7:45, could you not simply substitute x^2 for tx in y=tx? Why do we need to solve for x instead?
Me in precalc cp just learning about trig functions binging your videos: Wow, I wish I knew what was going on.
If you take any number for y, you can always take x=y^y!
because you will have y^y^y on bouth sides...
like 3 and 27 as shown in the graf-solution, you can do it with 5 and 3125 or any number
I hve learn new thing today, thanks
Carefully picked rational values lead to nice solutions. Say, anything of the form t = (a + 1) / a. Take a = 3. Then we have x = t^(1/(t - 1)) = ((a + 1) / a)^(1/(1/a)) = (a + 1)^a / a^a, and similarly y = (a + 1)^(a + 1) / a^(a + 1). So they look kind of messy, but plugging in numbers makes them much simpler. (2, 4), (9/4, 27/8), (64/27, 256/81), and so on. You could play around with other parametrized rational solutions that happen to work well and see what you get.
9:53 Why do you divide by t and then, after substituding, multiply both sides by t??😂😂
7:17 Should you ensure that x≠0?
Yes we want a video about pytagorian triples #yay
Yahya Hamid I second this
Yep
XaXuser it is easy man
Let k and n be any positive integer
A = k^2 + n^2
B = 2kn
C = k^2 - n^2
XaXuser what to proof here
Νικος Μανε what !? If we work in the set of positifs integers the your A will be greater than your C while the C is the hypotenuse?
7:58 Notice that when he divides by X, he asumes that X can't be ZERO, which makes a lot of sense since X is Real and if X= 0, the equation would be: 0^y = y ^0 and we all know that 0 to the power of any number (except zero) is 0 and every number to the power of 0 is 1, so we would get an absurd expression: 0 = 1 if X or Y is equal to zero
really nice video! but can you also solve x^2=e^x with it?
Philipp Huber parametric equations is used to solve implicit equations
Philipp Huber while you can still solve it using the lambert w function
x^2=e^x
x^2 e^(-x)=1
xe^(-1/2x)=±1
-1/2xe^(-1/2x)=±1/2
-1/2x=W(±1/2）
x=-2W(±1/2)
while W(-1/2) is complex
so the real solution of x is-2W(1/2)
Philipp Huber en.wikipedia.org/wiki/Lambert_W_function
無謂 I generalize it with this formula
x^p=b^x
x=-pW(-ln(b)/p)/ln(b)
無謂 ok this is a little too high for me
Nitpick at 7:25. You should note x can't be equal to 0 so taking the power of 1/x is valid. Implicitly y can't be equal to 0 either since y = tx.
I've recently had to solve sin(x)^cos(x)=cos(x)^sin(x) on my math exam 😀
Hmmm, x = pi/4 +2npi?
@@blackpenredpen yep :D
If you set x1. Setting t>1, you can prove there are no integer solutions by showing t^(1/(t-1)) only has an integer solution for t=2, which is EZPZ. Pretty cool : D
This guy is a sorcerer! A sorcerer I tell you!
Your second solution is nice for proving that 1^infinity is indeterminant. As t gets very large (approaches infinity), 1/(t-1) will approach 0 => x will approach 1, and t/(t-1) will will approach 1, which means y will approach infinity. You'll get a conclusion that 1^infinity = infinity, which is an alternative to the intuitive solution that 1^a = 1 for all a.
If you wanna respect Math, respect this man first. #YAY
You should do a video about Lambert W function
Finding complex solutions is boring ;) Try solving it in whole numbers ;)
Another fun thing to do would be to prove the identity you found (that `√3^√27 = √27^√3`) algebraically instead of depending on Wolfram ;> (because Wolfram can fool you, and it happened before, remember? :q )
Sci Twi yoooooo welcome back!!!!!!!
√3^√27=√27^√3
√3^(√27/√3)=√27
√3^√(27/3)=√27
√3^√9=√27
√3^3=√27
3^3=27
Are there other solutions ?
a^x=b^x=> a=b ?
Wrong if x=2 and a= -b0
Bro’ don’t forgot X & t are different to 0 (cause you divide by X & t)
this is obvious in case of x because if we allow it to be 0, then 0^y = y^0 => 0 = 1, a contradiction
The solution to this problem for natural numbers goes like this
x | y:
1^(1^1)
2^(2^2)
3^(3^3)
4^(4^4)
5^(5^5)
What a lovely pattern.
I don't understand the proof. You start with the assumption that y is a multiple of x ie y=tx. From there you generate real pairs as functions of t. But what if your initial assumption is not correct? What if you start with an alternate assumption: for example that y is an exponential power of x?
Nick Kravitz well in that example with the exponentiation, that is covered by that case, since y=x^t would mean y=x^t-1 * x, and x^t-1 is just some constant (I will write it c) so y = c x which is the assumption. The assumption is fine as long as we assume that both x and y are not 0. Since t=y/x will always satisfy y=tx.
When we say "multiple", we usually mean an integer multiple, but in this case we never specify that t has to be an integer. So in fact, we can always write y=tx, unless x is 0.
Nick Kravitz - he is not saying that y = tx for a _constant_ value of t. He is actually parametrizing y = f(t) and x = g(t) in such a way that f(t) = g(t)t.
@@mxlexrd yeah it feels like an odd choice of words
Este canal vale oro.
WHY IS THERE A FREKING HAIR IVE BEEN REMOVING IT FROM THE PAST 40 SECONDS
@@SaketRaman-jh6fw XDDDDDDDDDDDDDD