X € R /X#0

Thank you!!!

Nice

Manny Paul A skill you can’t help but learn when you teach lol

@@DarthAlphaTheGreat Xxx

@@amirnuriev9092 was

GreenMeansGO : )

LOL

How did it go?

It works with girls!

@@davidbrisbane7206 what age-group ?

Casey Roberson lol : )

They legendary.

@@15schaa They're*

@@RubyPiec shut

@@wilfriedsteinbach8700 up

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

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

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

@@sly317 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

@@sly317 of course. But that was the example in the video.

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

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.

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.

@@alvachan88 lol never seen anyone do that except this guy

Musicians rule: there is always an asian better, and younger than you. Whether you are asian or not.

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

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

Bryan Hart an ad?

I don't get it

Best would be to clear your acne...

looks fucky and here's sexy

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

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

Bautista Bauto98 thanks!!!!

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

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..

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.

: )

L1N3R1D3R Ahhhhhhhh loll!!! I didn't see that

Hahahaaa

Haha yep!!

Noticed that as well lmaoo

: )))))

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

: )

Charles David thank you!!!!

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

Well, there obviously are no (sane?) solutions for x=0 anyway, so...

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

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.

Thank you!!!

Thanks!!!

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.

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