This Video Will Make You Better At Algebra

How do you suppose you might solve this interesting equation x^y=y^x ? There are more than a few ways to do this! We will do it by introducing a parameter and solving for x and y in terms of it.
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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
x ^ y = y ^ x
#math #brithemathguy #algebra

[@BriTheMathGuy]

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[@axbs4863]

I always find it so weird how you can add more variables to make a problem less confusing Thank you for this insightful video and I’m really enjoying this series!

[@BriTheMathGuy]

Glad you liked it!

[@gol-r5836]

Ikr

[@aviancrane]

It's like getting a box and stuffing stuff inside it. The result is less to deal with. You just gotta make sure that if multiple boxes have the same label, they have the same contents so that you can treat them the same way.

[@createyourownfuture5410]

It's called parametrization.

[@kunalkashelani585]

It's not weird at all though. If you use it well, it can always help simplify things!

[@zone1035]

legit took me 40 seconds to get even more confused than i am during class

[@Private_Account101]

ong 😂

[@dukenukem9770]

This is one of the problems that I did with my son when I introduced him to the Lambert W function. I’ll have to add this alternative solution technique to that worksheet packet. Thanks for sharing!

[@General12th]

I was going to ask how it went with your son, but then I realized we already talked in one of Michael Penn's videos. I'm glad to see other people who also enjoy so many different math channels!

[@dukenukem9770]

@@General12th Yep! LOL That was me! I definitely subscribe to a lot of math channels... I figured that I was not alone, but it's always nice to encounter a fellow math geek!

[@Tom-vu1wr]

@@dukenukem9770 how old is ur son

[@dukenukem9770]

@@Tom-vu1wr He just turned 13. He’s been homeschooled in the subjects of math and physics since 2017. He transitioned to full time homeschooling during the pandemic closure. He was 12 when we covered the Lambert W function.

[@Tom-vu1wr]

@@dukenukem9770 lesgo. I always wondered what it would've been like to learn maths and physics from a young age in stead of starting basically at 16 with the basics. Maybe I'll do that with my kids too one day. Has he done any challenges or Olympiads yet?

[@michiell.2769]

You didn't include the solutions where y=x (when m=1, all pairs (x,x) are solutions except for x=0). Other than that, very interesting.

[@SolomonUcko]

In that case, m=1 would result in division by zero, which in this case means check the original equation

[@bloxrocks5179]

I think if you take the limit as x & y approach h you get h^h = h^h making h=0 (and by extension x=y=0) a valid solution, or at least its limit.

[@michiell.2769]

@@bloxrocks5179 If you set 0^0 equal to 1, then (x,y) = (0,0) is indeed a solution

[@otherodd]

@@michiell.2769 But you can’t just set 0^0 = 1, technically it’s undefined (although 0^0 = 0^0 should be still true)

[@michiell.2769]

@@otherodd Well, sometimes 0^0 is left undefined and sometimes it is set equal to 0^0 (because some theorems like the binomial theorem require 0^0 to be 1). From Wikipedia: "Zero to the power of zero, denoted by 0^0, is a mathematical expression with no agreed-upon value. The most common possibilities are 1 or leaving the expression undefined, with justifications existing for each, depending on context. In algebra and combinatorics, the generally agreed upon value is 0^0 = 1, whereas in mathematical analysis, the expression is sometimes left undefined."

[@berengriffiths656]

What's even more interesting is that this result still holds for irrational AND complex x and y (although solving for m to find y from a given x involves the use of the lambert w function). It's fascinating that a seemingly specific case at the beginning leads to *all* solutions! Great video!

[@JobBouwman]

All solutions? If think the set y=x are solutions as well.

[@Zevoxian]

@@JobBouwman that’s because we have in the exponent a divisor of m-1, so m=1 is the only special case not covered, which corresponds with x=y

[@daniel-kun6443]

at 0:40 to assume y=mx you will need v|u but nothing states that. Is the whole thing wrong?

[@InstigationMex95]

Quickly becoming one of my favorite math channels. Such clear and intuitive explanations

[@BriTheMathGuy]

Wow, thanks!

[@thegamingexpert156]

@@BriTheMathGuy I think you and MathAntics are the top two math channels.

[@BalaVerde]

lol clearly you don’t know how bad I am at math, you lost me 30 seconds in so…

[@yj_holic4590]

Never thought about it in terms of fractions! I only new 2^4= 4^2 = 16

[@jasnoor8-d-155]

Hey which editing app do you use ? I would really appreciate if you tell

[@BriTheMathGuy]

Hi there, thanks for watching and your interest! I'll say that I do not use Manim. I hope you'll forgive me and understand I don't feel comfortable sharing how I make my videos (at this time).

[@jasnoor8-d-155]

@@BriTheMathGuy understandable , have a great day !

[@ghostsmoke11]

this made me worse at algebra and increased my anxiety, thanks

[@jimmea6317]

I just thought “for all values where x=y”.

[@925NC]

Something doesn't add up here. First, your line of reasoning requires that x and y be natural numbers, but your result includes rationals as well. How can it be said that x^y represents some multiple of x when y=4/3? Also, your construction of x and y in terms of m excludes m=1, but y=1x is clearly a solution. I'm not disagreeing with your result, but there has to be a better approach than the one shown here. Still interesting nonetheless.

[@ezraguerrero2879]

Well, you can proceed as he did by ignoring the "multiple" talk. So long as x is not 0, there exists an m such that y=mx (take y/x). Then proceed as he did. As for m=1, this solution disappeared when he raised both sides to the 1/(m-1), so it is worth it to mention m=1 still works.

[@925NC]

True, but that begs the question are there other families of solutions such that, for example, y=mx^2 or y=m log(x), etc., does it not?

[@ezraguerrero2879]

@@925NC I guess you could try said substitutions, but I reckon you should get this same family. Like, we aren't assuming anything by making this substitution, so no solutions are lost. Think of it like this: Given a pair (x,y) that satisfies the equation, we find (x,y)=(m^{1/(m-1)}, m^{m/(m-1)}), where m = y/x. This is true for **all** solutions, so this gives a parametrization of **all** solutions. In particular, when you choose a specific value of m, what you are doing is selecting the pairs (x,y) such that y/x=m. Substituting y=mx^2 should probably give the same parametrization, although it may look different since m represents something else. For y=m log(x), you'd have to be careful that you are killing all solutions where x is non positive. Hopefully that made sense.

[@pawemarsza9515]

@@925NC there are two families of solutions. One is shown in the video, and is aquired from: x^(m-1)=m By taking (m-1)th root on both sides. But that step only works when m=/=1, as you can't take 0th root of something. So, the second family of solutions arises from the case m=1 Then the equation in question becomes: 1=1 which is true for every x (and y=x from our definition of m). So the second family of solutions is set of all pairs (x,x) for positixe x. Yet another, alone solution can be extracted at the very beginning, when we take x-th root of both sides of x^(xm)=(xm)^x Again: we can't take 0-th root, so we check x=0, and we get: 0^0=0^0 So it works, and y=m*0=0 Solution (0,0) can be then put into (x,x) family, so now x is nonnegative

[@JordanMetroidManiac]

It just means that not all solutions were actually found for x and y, just a family of them.

[@nice_mf_ngl]

I'm currently in 7th but i am preparing for the maths Olympiad in india (IOQM and INMO), your videos are super relevant and interesting, thanks for this video and please try INMO questions they are really interesting

[@aaravarora2009]

Oh Nice. I am also studying in 7th class from India and preparing for Olympiads like IMO and IOM. Yeah, his videos are really helpful.

[@rushilpatel7418]

good luck!

[@devd_rx]

Good luck bruv

[@meltedice-cream6499]

these videos cant be relevant u should focus one level up to say the least these are just for entertainment 😐

[@elizamaria1675]

@Emanuel Rob Și eu tot pentru olimpiada la mate în a 8a :)))

[@naivedyam2675]

Hey Bri. Here's a problem created by me. Do give it a try- If (dy/dx)^2 + y^2 = 4y, find derivative of dy/dx at y = 2 - root2. Condition is you can't use differentiation at any step.

[@ishaqhamza2729]

Solve the differential equation Then find the value of y at x=2-root2 Substitute in the diff-eq

[@naivedyam2675]

@@ishaqhamza2729 That equation isn't solvable. That's the point!

[@ishaqhamza2729]

@@naivedyam2675 well yeah, there ain't a unique solution...

[@naivedyam2675]

@@tddupaid Yes initial conditions matter. And you can't use differentiation at any step. How do you solve that eqn though?

[@michiell.2769]

You need more information (initial conditions). All functions y = 2 + 2 sin(x+C) (with C a real number) satisfy the differential equation (dy/dx)^2 + y^2 = 4y, so the derivative of dy/dx at x = 2-sqrt(2) is not uniquely determined. Edit: Ok, you changed 'at x = 2-sqrt(2)' into 'at y = 2-sqrt(2)'

[@lazaremoanang3116]

We have for example (√3,3) and (4,2). Ok let's watch the video.

[@Blaqjaqshellaq]

If m=1, x and y can be any identical pair of numbers!

[@mike1024.]

You missed that your solution assumes m is not equal to 1. When it is, that's one of the easy solutions where y=x, well when x isn't 0. The motivational thing you did at the beginning didn't really make a difference though. Since we're looking for numbers, sure any value of y can be a multiple of any value of x! It should even work for complex numbers.

[@tausifraza2506]

X = 2 & Y = 4 is also satisfies the equation Other possible combinations X = -2 & Y = -4 X = -2 & Y = 4 (and vice versa)

[@GordenRamsi20]

Indo?

[@literallylegendary6594]

(-2)^4 = 16 while 4^(-2) = 1/16, so not X = -2 & Y = 4 (and vice versa)

[@chunqiangli3636]

How can you be sure that X is linear with Y？ You can not take the relation of linearity for granted

[@Abedchess]

I think m is not a constant. M is also a variable, dependent on x and y. Y is not directly proportional to x (I don't fully understand his argument but this is my best interpretation) Y^x = x^y Y *(y^(x-1)) = x * x^(y-1) Y = x * (x^(y-1))/y^(x-1) Let m = (x^(y-1))/y^(x-1). So Y = mx, but m is also a variable, dependent on x and y. Y is not directly proportional to x.

[@joblox9199]

im understanding less than i did before

[@nishkarshyadav552]

Guess this video's not for me (10th grader)

[@BananaNik]

Bro I don't know if you already have it but a tiktok or youtube shorts account would blow up

[@Yotrek]

Tip when making math videos. Show *every* step, one step at a time.

[@hyuckra.7163]

i haven’t watched the video yet but i just thought “16” assume x = 2 y = 4 2^4 = 4^2 16 = 16

[@AmitSingh-sf5qp]

KZbin, please fix this video it has a dislike button . 💗 Loved this video on algebra . 🏵️

[@Dj0enderman3000]

Doesn´t make me any better at algebra... cuz I don´t really get the first 2 or 3 steps up to 0:40 XD My brain doesn´t like the idea that x y times is the same as something multiple of x (same for y of course), how can it when x times x times x is not something times x but x times x times x

[@fizisistguy]

I lost it when he said that some multiple of x is equal to another one of y. Can anybody explain?

[@IngTomT]

What about x=y?

[@Avinashkumar-zz8qi]

Mathematics never disappoints me! Mathematics is my only love 😍

[@AbhishekSingh-qn4bz]

BRILLIANT is absolutely Amazing..thnx

[@BriTheMathGuy]

Glad you like it!

[@WestExplainsBest]

Even the font you use looks sophisticated and mathematical. What font is that?

[@aniruddhvasishta8334]

I'm pretty sure it's just LaTeX math mode. I'm not sure what the font is actually named.

[@kaivamannam5636]

the LaTeX default math font is called Latin Modern or Latin Roman

[@benjaminmoreau3560]

Doesn't this also work for x = y ? Because this solution isn't included in yours

[@nimmanishanthreddy787]

Idk what's that advanced maths and I don't understand anything you said but it's true, when x is 2 and y is 4

[@manucitomx]

I love these very clear videos. Thank you for your hard work. (I do miss the ones where we actually saw you, though).

[@BriTheMathGuy]

Glad you like them!

[@Fire_Axus]

your feelings are irrational

[@aren1939]

I love complex problems and i love it even more when i can solve them with some help

[@mischa3039]

why tf does this video gets in my recommended after my final math exam..

[@UncreativUsername]

ummm idk if im dumb but im already lost as to how you got from some multiple of y is some multiple of x to y is some multiple of x.

[@zealous2835]

You know I really want to be an Astronomer when I’m older but after watching this video I don’t think I’ll ever be able to get a physics PHD lol

[@gioret998]

Hahaha same

[@fragile3169]

thanks, now i'm "algebros"

[@GordenRamsi20]

I didnt realize this was a new video

[@BriTheMathGuy]

Thanks for watching and have a great day!

[@alecbuckley8314]

couldn't...couldn't 2 and 4 work though?

[@TrimutiusToo]

You missed y=x solutions which is when m=1 which you forgot to account for

[@DeJay7]

when m=1 we divide by 0, which in this case means if x=y it is true for all x, y

[@TrimutiusToo]

@@DeJay7 we do not divide by 0, because he divided by (m-1) without checking whether it can be 0, when it actually can, so for (m-1)=0 extra work needs to be done

[@cosimobaldi03]

@@TrimutiusToo yeah, probably along the way he made an operation that assumed that m cannot be 1. Also if you look at the graph of all the solutions there are two distinct groups: the one he found and the one with all the x=y...so it makes sense that a simple function that gives solutions cannot find them all, just like x=y doesn't find all of them

[@TrimutiusToo]

@@cosimobaldi03 yep he divided by (m-1) along the way without checking if it can be 0

[@TimeFadesMemoryLasts]

Wow, I didn't expect to get an infinite amount of solutions. I was expecting 1, 2, or 5 solutions or whatever...

[@flacsomtodosclas2165]

1^2≠2^1

[@foji-video]

you actually get to that solution without involving going through the "let's consider this as if it was a natural exponent" part, with may seem non-rigorous. The reasoning holds by just going with factorization: x^y = y^x --> (x^(y-1)) x = (y^(x-1)) y . Now you can call x^(y-1) as your u and y^(x-1) as your v, and you get that m. Of course m depends on x and y, but as you've shown you can actually get the local "inverse" of the m(x,y) function and get y(m) and x(m)

[@FleuveAlphee]

The natural number (unstated, except for the - added later? - note) assumption is even more non-rigorous than it seems. If x is a natural number, what is 1/x? And what about "generalizing" to any real number? Where is it done? And justified?

[@foji-video]

@@FleuveAlphee yeah, that's also a good point I didnt consider. Nevertheless, you can get out of it easly. It's actually a decent example of "sometimes non-rigorous and somewhat fuzzy reasoning gets you to good result". But mathematics is also full of the opposite case

[@Surya-cc7it]

4^2 = 2^4 Hence, x = 2 and y = 4. Hence solved.

[@tmoneytheshooter4208]

I'm new to the channel

[@rubenvela44]

I love you

[@coolj4334]

Let's say that x=1 and y=2 1^2=1 2^1=2 1≠2

[@SeegalMasterPlayz]

x^y = y^x is LITERAL Easy!!! 2^4 = 4^2!!

[@Hippolyte_Pequeux]

3:35 I don't get why y is a multiple of x... u · x = v · y doesn't imply that y = m · x

[@salisghaier3074]

yes i think so to , the only way it is true is when y and u and prime to one another

[@ShadowPhoenix4798]

X and Y could also be equal

[@kaustabhchakraborty4721]

Mr Bri could you plz see to the problem, it's something I could not proove nor find any on the net. Proove that a1^3+a2^3+a3^3+...an^3=b1^3+b2^3+b3^3....bn^3 iff a1=b1,a2=b2,...an=bn I don't know even if can be prooved or not. Even if the above equation could be proved for a power greater than three then also my job would be done. Plz see if you could help, plz.

[@fullfungo]

It’s not true. 1^3+12^3=9^3+10^3.

[@mark7142]

search taxicab numbers on google, you can find many counter examples

[@kaustabhchakraborty4721]

@@mark7142 so two questions, first - is my equation not true even for for n>=3 and if my equation is true for any power greater than three.

[@anderskallberg7969]

Your question: iff a1=b1,a2=b2,...an=bn... then is a1^3+a2^3+a3^3+...an^3=b1^3+b2^3+b3^3....bn^3 ? How can it not be true? If a_i = b_i, then each term on the left side has an equivalent term on the right hand side, no matter the exponent on the series. if n=4 => a1^3+a2^3+a3^3+a4^3 = b1^3+b2^3+b3^3+b4^3 = a1^3+b2^3+b3^3+b4^3 [a1=b1] = a1^3+a2^3+b3^3+b4^3 [a2=b2] = a1^3+a2^3+a3^3+b4^3 [a3=b3] = a1^3+a2^3+a3^3+a4^3 [a4=b4] Which is now identical to the left hand side. True no matter the exponent. Seems to easy, did I miss something?

[@kaustabhchakraborty4721]

@@anderskallberg7969 no no, all the permutations of a and b you have written are the same as the original, no different set of numbers were found which equate to the original equation other than a1,a2,...a3

[@arqade3452]

Real formula : x^y = y^x + (x-y)

[@lofivibez182]

How in any way is this simple?

[@nicolastorres147]

2:20 we can actually graph the solution curve r(m) = (x(m), y(m)) in geogebra

[@zombeaver69]

algegraphy*

[@diogenesfeuerbachnietzsche8848]

Now my brain hurts

[@khangphuc778]

me: 2^4 = 4^2

[@medaminehannoune4280]

When we did x^y =x.x.x… we actually supposed that y is an integer whereas we found a none integer as a solution Absurd

[@ozzycodm6743]

I'm in 7th grade.... It still confuses me and Im in trouble bc the deadline of my modules are tomorrow at 11:59 pm and Im cramming coz I have to do 3 subjects which is Maths TLE or Technology Livelihood (I probably even spelled that wrong) and Education. Which is basically Robotics.

[@uforob5601]

how this relates to all solutions with y=x with x > 0?

[@cosimobaldi03]

since m is y/x, for this class of solutions m is constant and gas value 1. If we plug m=1 into the expression at the end we get x=y=1^inf, which Is undetermined... We could interpret this as saying that no matter the value we give to 1^inf, it is a solution of the equation

[@Zettabyte420]

This video: (4^(1/3))^(4^(4/3)) = (4^(4/3))^(4^(1/3)) Me, an intellectual: 2^4 = 4^2

[@AlEx-mj8ol]

This solution has a major flaw - wince we assume our x and y can only be positive numbers (otherwise we cannot use non natural numbers as posers), but there are plenty of solutions that we miss x=y in Z^-. (problem of finding answers in R^- is even more comlex)

[@AndreyGoryainov-k7o]

This can be solved much easier with the fundamental theorem of aritmetic

[@nicolastorres147]

How?

[@zakaria3622]

bro u lost me at 1:09

[@Deeer69420]

Can someone list all the answers of m

[@Deeer69420]

I meant solutions

[@newnoticedthings4963]

Yes brilliant is a fun and interactive way to learn but again one more helpful website that helps sharpening mind memory with fee😒😒

[@Amar-lv1yw]

I think you made an mistake: x * x * x * x.... does not equal u * x. u * x would actually be: x + x + x.... u times. Or did I misunderstand something?

[@cosimobaldi03]

x * x * x... Is definetly a multiple of x, so It can be expressed as x*u. In particular x^y = x * x^(y-1)

[@Amar-lv1yw]

@@cosimobaldi03 i think i know what you mean. It wasnt immidietly clear to me. Thanks

[@olejrgenbrnner4708]

@@cosimobaldi03 The video kinda presents it as the two equations are equivalent, which they're clearly not as there's a lot more u's than x^(y-1) when you assume natural numbers. I guess by "luck" the generalization of exponents make this work regardless, but it's far from obvious?

[@olejrgenbrnner4708]

Also, the fact that u is assumed to be a multiple of x does not seem to be used in the derivation.

[@MessiBoT]

2:48 The most important sentence for me, I really needed it.

[@JobBouwman]

I didn't understand your motivation that y is a multiple of x. It just doesn't follow from your explanation. I know that the substitution y = rx is useful, but you confused me. I believe you haven't clarified this.

[@SuperYoonHo]

nice

[@MDSiam-fj4nb]

If you assume that, x = 5 & y = 3, the result would 125 = 243. And I don't think that's correct at all.

[@luukipuuk3537]

I'm not good enough at math to understand any of this. But it's ok. I'll check again when I'm not extremely hungover!

[@batteryjuicy4231]

So m is he patameter of the second line that intersects the y^x =x^y line. So if we subtract the 2 equation you found the end, we get the equation of the second line. Did i understand it correctly?

[@matejprokop7923]

I did solve it using graphs. this thing is always gonna be true as long as x = y

[@supertlp1062]

That is a graphical solution. If x=y it is very easy to see that x^x=x^x and y^y=y^y. You can see it with your eyes

[@olanmills64]

But that's not the only solution. Also, you don't really need a graph to see that x=y would be a solution to x^y=y^x

[@supertlp1062]

@@olanmills64 graphical referred to your eyes

[@olanmills64]

@@supertlp1062 I intended my post as a reply to Matěj specifically

[@thedancingscientist8180]

hahahahhaha this is absolutely cool, elegantly mind-boggling to say the least. Just how math sometimes makes sense when it doesn't make sense

[@Fire_Axus]

but the solution for the variable y involves a non-elementary function...

[@jmadratz]

Infinite number of solutions

[@prazum]

2^4 and 4^2

[@kubekson7786]

I don't understand. Imagine x=2 and y=3, then its all wrong

[@Abedchess]

? 2^3 is not equal 3^2,

[@Abedchess]

Which part of the method confusing?

[@Abedchess]

The question is asking you to find solutions to the equation x^y = y^x An example of a solution is x =1, y =1 Or x=2 and y= 4

[@rejuksss]

why not just raise both sides of the equation with 1/x so that we get a function y=x^(y/x)?

[@WillRlox]

WHERE IS MY GRAVITTY FALLS VIDEO YOU KNOW YOU CANT DO CLICKBAIT RIGHT

[@ralphparker]

So M is any real number except 1? Does M include any imaginary numbers?

[@abiyhailu2430]

How about X=2 and Y=4. I get it by my own means. I will show u if u are interested .

[@prashjobs2795]

It doesn't work to x =3 and y =2

[@뀽-Rbd]

School should teach this things to students rather than making students to solve billion math problems. - korean student -

[@justinkianaalfredo6843]

how stupid i am i thought the answer is simply x = 2 and y = 2 because both equals 4

[@GEORGIOSMGEORGIADIS4]

Excellent!! 💯💪

[@BriTheMathGuy]

Glad you liked it!

[@llo5673]

well if i would given the m i could find the x and y in for the original notation of the problem. So i cant see how. this manipulation is useful

[@azbeatz8451]

put x=2 and y=3 and your equation proves wrong....

[@_Rainbooow]

fun fact: y^x is never equal to x^y (it is possible if y = x)

[@ATKPerc]

I don’t understand why we’re putting 1 over x?

[@StephenwithaPH]

I don't understand wtf did you talking about, but ok nice video

[@calebdavies526]

I think I'm actually worse at algebra now

[@josephyoung6749]

I enoy these videos but they feel like commercials with a little bit of info on math at the beginning

[@Isakovsck]

They're both 1 *leaves the room* *gigachad gif*

[@issamelkonian963]

no you didnt solve it , you cant put m=1 , but if m=1 , y=x , x^x=x^x

[@paolot9774]

A question: the approach x^y and y^x means x multiplied by itself y times and y multiplied by itself x times does not work only if the number of these times is integer?