Math Olympiad Question | Equation Solving | You should learn this trick

How to solve this equation? By using this trick, you can deal it quickly!
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[@mathwindow]

If you have any suggestions or questions on math, comment as a reply! ❤😊

[@SiwakhileVusimuzi]

Am having difficulties with this x^x^27=77 😢

[@Taric25]

You forgot the imaginary solutions ±(⁴√8)i.

[@BongiMojapelo]

So why did the person say( x^ x4)4 where did the get the 4 in the first solving stage

[@spymadmax584]

I don't know but your skill of writing clean and symmetrical brackets amazes me, cuz all I get is like one small and one giant bracket whenever I try

[@shubhkapoor1940]

Unique way of simping Noiceee

[@Snoopyguys]

Bc

[@raid6n529]

@@shubhkapoor1940 what is simping?

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@KSY42]

Хороший маркер и всё получится.

[@hydraslair4723]

You can avoid having to think about tricks if you define u = x^4. Then you get u^(u/4) = 64 which calls for raising both sides to the fourth power. You end up with the same equation

[@alexandrenouhet4669]

Can you write the details pls for my little brain ?

[@giobur]

@@alexandrenouhet4669 So because of the properties of powers, ((x)^x)^4 = (x)^x⋅4 = (x)^4 ⋅x = (x)^4^x. So we ca substitute u to x^4 an get (u)^x. Now to express that other x as a function of u as well, we can observe that x = √(√(u)), and that doing the square root of something is the same as elevating tha something to the 1/2th power, so we have x = ((u)^1/2)^1/2 = (u)^1/2⋅1/2 = (u)^1/4. So if we substitute this to x in the equation (u)^x we get ((u)^u^1/4), that like before is equal to (u)^u⋅1/4 = u^u/4. So back to the original problem you now have u^u/4 = 64, and you can elevate both to the fourth power and get u^u = 64^4. After this the explanation in the video will get you to the solution. Hope this cleared it up!

[@voyag1473]

Substitution is the universal solution for this type of solution because when the next time u see something more complicated u may probably not be able to rearrange it into 8^8

[@alexandrenouhet4669]

@@giobur thanks a lot mate !

[@user-gy7zo1eq5z]

Yeah it make much more sense

[@newshhh]

Deducting from x^x = 8^8 that x = 8 is a typical mistake in math. It only shows x = 8 is one of the potential solutions, but doesn't rule out other solutions if any.

[@andrewclark2503]

Isn't the function f(x) = x^x (i) increasing in x when x > 1 and (ii) satisfying f(x) < 1 for x < 1? So there is exactly one solution to f(x) = y for any y >= 1.

[@leweeb949]

@@andrewclark2503yulkMK yuko

[@FitzTomBlaireau]

@@newshhh Thanks for supplying this insight, it's much more satisfying than this vid (which I clicked on expecting a method)

[@dudewaldo4]

@@newshhh This is like saying that any time you solve a linear equation you have to include a proof that linear functions are monotonic and continuous. That is silly

[@lukandrate9866]

She does it because in other case you'll end up trying to solve a transcendental equation with non-elementary solutions while trying to seem smart instead of just solving for the solutions for which you can solve

[@musiquinhaslegais4097]

One little detail: it is useful to note that the function (x^4)^(x^4) is monotone for x>1 and is 1)

[@theofigueiredodamasceno5601]

Não entendi o que você quis dizer com monótono pra x>0, explica pfpf.

[@musiquinhaslegais4097]

@@theofigueiredodamasceno5601monótono quer dizer que é crescente ou decrescente, não pode oscilar. Nesse caso, a função vai sempre crescer a partir de x=0 (na verdade a partir de x=1, tinha erro na resposta original), então depois que ela passar pelo valor desejado, não volta mais, então só vai assumir o valor essa vez pra x>1. E se 0

[@theofigueiredodamasceno5601]

@@musiquinhaslegais4097 Entendi! Muito obrigado.

[@nikitapickf8489]

This function is not monotone for x>0

[@musiquinhaslegais4097]

@@nikitapickf8489 I think you're right. Correcting it: if |x|=1, but since the function iis monotone for x>1 and for x

[@artkirakosyan2633]

In this particular case you may be lucky and it will be the solution. But you did two mistakes already, first when you raise something to a degree of a even number then you can creat more roots then it really has, so x4 range needs to be discussed. Second when you just put x equals to eight it may be not the only solution. Or maybe it is but it needs to be addressed. I wouldn’t recommend learning math this way. It is better to understand what you are doing and never solve the equation then writing lots of bullcrap without understanding.

[@kobalt4083]

Taking the fourth power gets us (x^(x^4))^4 = 64^4. We can arrange it into the form a^a by noting a^mn=a^nm. So (x^4)^(x^4)=64^4=(8^2)^4=8^8. Therefore x^4=8, which we can write as the only solution since the function y=x^x monotonically increases when x>1/e, where e is Euler's number, to be specific. Now we can solve the quartic x^4=8 for the four solutions by factoring or square rooting the equation two times. Factoring: Bring 8 to the other side, then rewrite the difference of squares -> x^4-8=0 -> (x^2)^2-(sqrt(8))^2=0 -> (x^2+sqrt(8))(x^2-sqrt(8))=0. So x^2+sqrt(8)=0 or x^2-sqrt(8)=0. The solutions to this are x=+/-4th root of 8 and x=+/-4th root of 8 * i. Square Rooting: Taking the square root two times gets us x=+/-sqrt(+/-sqrt(8)) -> x=+/-4th root of 8, +/-4th root of 8 * i.

[@Zwerggoldhamster]

I feel like you showed nicely that ±8^(1/4) are solutions, but you didn't show, they are the only ones. Imagine a modified version of x^(x^4)=1, then 0 and 1 would both be solutions of the equation (but -1 wouldn't).

[@jyotismoykalita]

In your modified version x^(x^4) = 1, If we put x = 0, then 0^(0^4) = 0^0, which, is undefined and not equal to 1. So 0 wont be a solution to this equation.

[@brunocombelles59]

it's the transition from the green equation to the red equation that needs more attention. If A to the power of A is equal to B to the power of B, it does not follow that A = B as x to the power of x is not injective

[@stephenholt4670]

@@jyotismoykalita 0^0 is not "undefined", it is 1. x^0 is equal 1 for any x you choose, positive, negative or zero.

[@riggsmarkham922]

@@stephenholt4670 The problem is much more complicated that you think, and the true answer is that there is no universally agreed-upon value. It’s ambiguous as to whether it equals 1 or whether it’s undefined, and different parts of mathematics disagree on what it equals. In algebra, people tend to say that 0^0=1, but it’s important to note that that isn’t like an incontrovertible fact of the universe. It’s a weird, ambiguous concept, and people just decide to use one of them because it makes other things convenient. There’s a whole Wikipedia page about it: en.wikipedia.org/wiki/Zero_to_the_power_of_zero

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@jim2376]

The precision and readability of her writing are amazing. And multicolored for a bonus!

[@mathbook1993]

kzbin.info/www/bejne/aXqnkI2kpsiBgNU

[@errornotfound_1972]

*_dễ quá. bài toán này cách đây 17 năm tôi đã làm rất thành thạo. và bây giờ tôi vẫn còn nhớ rõ công thức của nó. tôi đã nhìn và nghĩ ngay ra đáp án. tôi thích môn toán có dạng mũ số, lượng giác, tích phân, giới hạn (lim). trong một kỳ nghỉ hè ngắn tôi đã ngồi làm hơn 1000 bài toán về các chủ đề như thế này. tôi thực sự đam mê với nó. tôi yêu môn toán hơn bất kỳ môn học nào khác. nó chiếm trọn thời gian của tôi mỗi ngày khi tôi còn học ở phổ thông_*

[@timeonly1401]

For these towers of power one needs to remember that a^b^c = a^(b^c). Since the righthand-side (RHS) is 64 = 2^6, it's not unreasonable to suppose that x is also a power of 2. Let x = 2^a. Using guess-and-check on 'a', with target for the LHS of 2^6 = 64: When a=0, x = 2^0 = 1: 1^(1^4) = 1^1 = 1 = 2^0 ≠ 2^6 (a is too small) When a=1, x = 2^1 = 2: 2^(2^4) = 2^16 ≠ 2^6 (a is too large) When a=1/2, x = 2^(1/2): [2^(1/2)]^{[2^(1/2)]^4]} = [2^(1/2)]^(2^2) = [2^(1/2)]^4 = 2^2 ≠ 2^6 (a is too small) When a = 3/4, x = 2^(3/4): [2^(3/4)]^{[2^(3/4)]^4]} = [2^(3/4)]^(2^3) = [2^(3/4)]^8 = 2^6 = 2^6, checks!! So a=3/4, and x=2^(3/4)= ∜(2^3)=∜8. Done! Took 10x longer to type & explain than to do this in my head!! (I was just looking for the real solutions.. have to think more about the -∜8 and i∜8 & -i∜8. Just checked WolframAlpha.com and they all give 64 when put into expression x^x^4. Who knew? ;-)

[@OrenLikes]

Very nice! Could go further and say that "x=±2^(3/4)" since "x^4=8=2^3" and rooting a number is dividing its exponent (not sure I used the correct English and Mathematical words/terms). Third root of 27 is "27^(1/3)" because "27=3^3" and third root of 27=27^(1/3)=3^3^(1/3)=3^(3/3)=3^1=3.

[@MikeN1811]

Уравнение имеет только один положительный корень 8^(1/4), так как второй отрицательный и не входит в область определения функции.

[@user-py6mf9yx2b]

А почему не входит? Даже если икс отрицательный, то парная степень минус убирает

[@MikeN1811]

@@user-py6mf9yx2b Рассмотрим функцию у = (x) ^1/2, она определена только для x>=0, для действительных значений x. Если показатель степени не 1/2, а также является функцией от х, то также будет ограничение на х, кроме того, ещё добавляется, что х не равен 0, так как не определено 0 в отрицательной степени.

[@toly1961]

@@MikeN1811 Подставьте -8^0.25 в условие. Получите тождество. А то, что не стыкуется с промежуточными выкладками, проблема этих выкладок, а не результата.

[@user-wz6ec2vo1b]

по определению показательной функции основание строго больше нуля

[@ArloLipof]

The function f: x → x^x = exp(x • ln(x)) is continuous on ]0;+infty[, strictly decreasing on ]0;1/e[ and strictly increasing on ]1/e;+infty[. So the injectivity of f, i.e. « f(x^4) = f(8) => x^4 = 8 » is only true iff x^4 > 1/e, i.e. x < -1/e^1/4 or x > 1/e^1/4 (which luckily works in this case).

[@Commander.and.Chief.Killua]

Try (+or-square root of 2square root of 2)^8

[@ArloLipof]

@@Commander.and.Chief.Killua Please only comment if you understand the needed concepts of mathematics. What do you mean?

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@carstenmeyer7786]

@@ArloLipof There may be an error concerning the range where *f* is injective, provided we only allow *x > 0.* The range where *f* is injective should be *x ∈ [1; ∞) ∪ {1/e},* as the image of *f* satisfies *f( (0; 1/e) ) = f( (1/e; 1) ) = ( e^{ -1/e }; 1)* Of course, the remaining argument still holds given *x^4 = 8 > 1.*

[@VeteranVandal]

2^(3/4) ? Ok, I got it, and yeah, the negative root also works. I used logarithms, but only because I didn't find a way to write it symmetrically as you did in your solution. I tried to write the exponential in other forms, I just didn't find the one I needed. Of course, with log you could do in your head if you used base 2. Otherwise you'd need to used a few log properties to get to the same conclusion, and that without a pencil is slightly more challenging. The complex solution, tho, is definitely more complicated.

[@obvioustruth]

Dude!!! He fucked it up! He got it wrong. He made mistake between first green line and first blue line. He assumed that (a^b)^c = a^b^c. That's wrong!

[@tttm4rt1n49]

@@obvioustruth he didn't

[@ayushmanchakraborty8838]

@@obvioustruth it's not wrong, go to junior school and learn about exponents, bloody illiterates

[@learf6613]

@@obvioustruth true that they are not equal but there wasnt a case of a^b^c = (a^b)^c, its just simply a^b^c times 4

[@jbrady1725]

I was trying to guess at it too, and eventually ended up at 4th root of 8, which is same as you.

[@Balila_balbal_loki]

There is are two missed solutions. i*(forth root of 8) and -i*(forth root of 8). The reason i'm including imaginary numbers is because this result ends up being harmonic.

[@michaeledwards2251]

64 is 2 power 6. 4th root each side. x power x = 2 power 3/2. Considering the logs, x is simply the square root of the power of 2 or 2 power 3/4.

[@oahuhawaii2141]

You aren't exponentiating properly: x^x^4 is x^(x^4), and not (x^x)^4. Thus: x^x^4 = 64 = 2^6 x^(x^4/4) = 2^(6/4) = 2^(3/2) At this point, the next steps in simplifying the equation aren't clear.

[@phuocvlog]

we just imply x^4=8 from (x^4)^(x^4) =8^8 when function y=x^x is monotone

[@duke6585]

As a 7th grader, I can say with confidence I’m ready for algebra 7836. Thanks for the quick lessons!

[@mangouschase]

i'm sorry but you are not, this things are just 10th grade (if i did the conversion correctly to american)

[@OSU23901]

@@mangouschase probably not the right conversion, I’m in 10th grade and this is a little bit ahead of what we do. Maybe 11th-college

[@dudewaldo4]

lmao you go kid

[@comptech5240]

@@mangouschase in all seriousness, someone who has learned exponents nicely in 7th grade will be able to do it easily in 8th grade

[@bmbelko]

I appreciate the step by step demonstration for a refresher.

[@user-ej2eh1zy2q]

The 8^(1/4) is ~~ 1,681739, which X^X^4 results in ~~ 33,02252 and not 64. What i am doing wrong?

[@oahuhawaii2141]

I did a few calculations to see how you got your numbers. You calculated 8^(1/4) correctly as 1,681792830507429..., but wrote the rounded value wrong in your comment. You intended to write 1,681793, but wrote 1,681739. This typo did not affect your next calculations. With your correct x, you calculated (x^x)^4 instead of x^(x^4). That's how you got 33,02252407144914... instead of 64.

[@b-penajohneric151]

so we can get 4√8 (1.68) and we can recheck the equation, actually it is 64 cause you cancel the 4√ on 2nd exponent against the 3rd exponent (which is 4) so you can get (4√8)⁸, and after that the certain rule of exponent applied in square roots against exponents so possibly cancel the 4√ and the 8 will reduced into 2 so 8² = 64

[@mathbook1993]

kzbin.info/www/bejne/aXqnkI2kpsiBgNU

[@joshmckinney6034]

Beautiful handwriting and beautiful explanation!

[@GDyoutube2022]

There is maybe a “simpler” solution involving smaller numbers. 64 is 2^6 so let’s simplify both sides and obtain x^x=2^(3/2). The fact one side is a power of 2 tells us that also the other side, hence x, can be expressed as a power of 2. Hence, we can solve for (2^y)^(2^y)=(2^(3/2))^(2^0). Then, we can just “redistribute” the sum of the exponents 3/2 and 0 into 2 equal halves, that is into 3/4 and 3/4. So, since y=3/4, then x is 2^(3/4).

[@mathbook1993]

kzbin.info/www/bejne/aXqnkI2kpsiBgNU

[@oahuhawaii2141]

You manipulated your exponents incorrectly: x^x^4 = 2^6 The 4th root is: x^((x^4)/4) = 2^(6/4) = 2^(3/2) Note that x^x^4 is x^(x^4), and not (x^x)^4 . You commit a similar mistake a few steps later.

[@GDyoutube2022]

@@oahuhawaii2141 I think you are not wrong, but I am not wrong either. :) Let me explain - we are starting from different interpretations of the initial equation x^x^4. I went for the "coding" interpretation where any software (e.g. MS Excel) operates the formula consecutively as (x^x)^x, while you went for the classical interpretation of x^(x^x). Let me finally argue that even with classical notation the same "simmetry trick" I propose can be leveraged.

[@oahuhawaii2141]

@@GDyoutube2022: Excel is full of bugs. Put extra parentheses to coerce it to do the right thing. BTW, Excel indicates 1900 is a leap year, yet it's well known that the Gregorian calendar makes century years non-leap years unless it's divisible by 400. This is the kind of bug you want to redefine long-established rules, such as those we use in math? It seems Microsoft can change all the math, science, and engineering disciplines by not fixing its many bugs.

[@str8l1ne44]

Daamn That was smooth and clear Well played mate

[@RileyRampant]

your handwriting is beautiful. Its a joy just to see you write these expressions.

[@sangjeonglee4182]

ln(x^(x^4)) = 6ln2 x^4 ln(x) = 6 ln2 here, x^4 = e^(4lnx) so, e^(4lnx) lnx = 6ln2 and 4lnx * e^(4lnx) = 24 ln2 Lambert W function, 4lnx = W(24ln2) x = e^(1/4*W(24ln2)) is about 1.68179

[@oahuhawaii2141]

x^x^4 = 64 = 8^2 (x^x^4)^4 = (8^2)^4 x^(4*x^4) = 8^(2*4) (x^4)^(x^4) = 8^8 I'm not going to use the Lambert function to find the many complex solutions of n^n = c because, with n = x^4, I would have to solve for x, which is the 4th root of each complex result, times the four 4th roots of 1. So, I'll just equate the bases to deal with the real solution of n^n = c and continue from there: x^4 = 8 = 2^3 x = 2^(3/4)*i^z, for z = 0,1,2,3 Note that i^z for z = 0,1,2,3 yields the 4 fourth roots of 1: 1, i, -1, -i .

[@victorguilherme7955]

I just found your channel right now and loved it. Very helpful! Thanks from Brazil!

[@mathbook1993]

kzbin.info/www/bejne/aXqnkI2kpsiBgNU

[@sourivore]

If x=7 then x-x = 8-8 BUT x is not equal to 8... You have to explain why if X^X = 8^8 then X = 8... (The reason is function x^x is an increasing function but you have to prove it too..)

[@oahuhawaii2141]

Lambert function shows that there are an infinite number of complex solutions.

[@hieuduong7447]

Math is interesting, but it can be very tough and time consuming. Solving maths is like doing an artwork. It’s logical but atheistic at the same time ❤

[@mathbook1993]

kzbin.info/www/bejne/aXqnkI2kpsiBgNU

[@froreyfire]

Artistic. :-)

[@froreyfire]

Artistic. :-)

[@varoonnone7159]

​@@froreyfire Maths doesn't have a god

[@froreyfire]

@@varoonnone7159 I do believe that God even created maths. But regardless, the word "atheistic" doesn't make sense in the OP's comment, so he probably meant "artistic".

[@tmp3477]

So, axing the exponent of your ex, gives your eggs in hex. Very clear.

[@antilex07]

Ну тут понятно сразу было, что нужно поиграть со степенями двойки. 2^1 - много; 2^1/2 - мало. Берём среднее 2^3/4, проверяем... И бинго!

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@user-im3pf1fs4p]

точно, влепить в уравнение не 64, а 65 или 63, накувыркалась бы с корнями различных степеней, это не железный алгоритм, из носа выковырянный

[@user-mh6ke4zs4c]

Я почти так и сделал. На задачу ушла пара минут. Только я сначала перевел 64 в 2^6=(2^3/4)^8=(2^3/4)^(2^3/4)^4. Чтобы получилась левая часть. Ну, и в итоге x=2^2/3.

[@Musabll78]

CommodoreC64=x=11=3=32*2=64

[@darkknight2414]

I dont know why I get these videos recommended at 3am and it leads me to try to solve this at night🙄

[@Anonymous.s.r]

can we solve it by taking log base8 on both side

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@oahuhawaii2141]

Well, you didn't try it because that doesn't work out.

[@tgx3529]

Yes, but I had similar Idea. x^4 logx=log64 (1/4) y log y= log64 where y=x^4, then log64=log(4^3) ylogy=4*4*log(4^2)

[@philippeb.8677]

Oui, on retrouve fréquemment cet exercice sur youtube. Le problème, c'est que la solution proposée, à savoir x = 8^(1/4) ≈ 1.681792830507 est mathématiquement inexacte, puisqu'elle ne vérifie pas l'équation: x^x^4 = 64 En effet: (1.681792830507)^(1.681792830507)^4 ≈ 33.022524071362993 ≠ 64 En revanche, la fonction de Lambert permet de résoudre ce problème: x^x^4 = 64 x^x = 64^(1/4) = 2√2 Ln (x^x) = xln(x) = ln(x).exp(lnx) = ln(2√2) Fonction de Lambert: ==> Lnx = W(ln(2√2)) x = exp(W(ln(2√2)) = ln(2√2)/W(ln(2√2) ≈ 1.7884541573524 Vérification: (1.7884541573524)^(1.7884541573524)^4 ≈ 63.99999999999648696 ≈ 64

[@tanvirahmed1033]

Outstanding and excellent

[@strayfox6835]

Good simple solution, but I found 4 roots: x^4-8=0 (x^2 - eigth root of 8)(x^2 + eigth root of 8) = 0 x_1,2 = ± (sixteenth root of 8), x_3,4 = ± (sixteenth root of 8) i

[@hedwig7526]

The perfection of her hand writing amazed me

[@panosdiamadopoulos1682]

W lambert function and is solvable in 5 lines without literally any thinking

[@Musabll78]

W^M Faster!

[@syberrus]

-8^1/4 cant be a solution because when raising to a real power (non-integer), the base must be positive by definition of raising to a real power.

[@oahuhawaii2141]

You have precedence issues. You wrote -8^1/4, which boils NK down as: -(8^1)/4 = -8/4 = -2 . She has -8^(1/4), which has a positive base: -(8^(1/4)) .

[@alittax]

Thank you, that was very satisfying to watch! Nice work!

[@shlokee]

if we use differentiation (or operator and log) it'll be a lil different

[@arjunsanap378]

Take log will be easy I think so

[@serhiislobodianiuk776]

The main idea is good but you are not allowed to put any negative value of x into the formula in the left side if you are solving the equation in reals. The two variable function f(x, y) = x^y is defined for x>0, y - any real. So you should have stopped on 2^0.75

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@Azif45628]

Since x^4 = x×x×x×x = x×(x^3) x^x^4 = 64 (x^x)^ x^3 = 64 (x^x)^ x^3 = 2^6 (x^x)^ x^3 = (2^2)^3 (x^x)^(x^3)2 = (2^2)^(2)×3 By equating powers 2(x^3)=3

[@JiminatorPV]

@@Azif45628 you asumed that x=2 by equating exponents

[@Azif45628]

@@JiminatorPV in such case powers couldn't equate

[@JiminatorPV]

@@Azif45628 that's what i mean, you can't equate powers here

[@altSt0rm]

There are an infinite number of solutions to this if you consider the complex plane as well 8^(1/4) * cos ( 2*N*pi / 4) + j* 8^(1/4)*sin ( 2*N*pi / 4) for N any integer. Solution in the video is a special case of N=0

[@eliremingtone4654]

NO если в задаче не указан тип переменной то он считается вещественным и не может иметь комплексное значение

[@lukandrate9866]

@@eliremingtone4654 "if you consider the complex plane as well" - не, не видели

[@eliremingtone4654]

@@lukandrate9866Does the problem mention the complex plane? Есть простое правило -- если ничего не указано числа считаются вещественными

[@oahuhawaii2141]

You only have 4 solutions: 2^(3/4)*i^k, where k = 0,1,2,3. That is really 2^(3/4) times the four 4th roots of 1: 1, i, -1, -i . You forgot that the sin(a) and cos(a) cycles every 2*π to repeat their values again and again. The narrator gave the 2 real solutions and missed the 2 imaginary solutions. If you use the Lambert function to solve n^n = 8^8, you'll have an infinite set of complex solutions for n. For each n, you should take the 4th root and multiply by each of the four 4th roots of 1.

[@kobalt4083]

For x^4=8, the only complex solutions are x=+/-4th root of 8 * i.

[@antoniusnies-komponistpian2172]

Though I'm not convinced yet that a=b is the only solution for a^a=b^b

[@oahuhawaii2141]

Look up the Lambert function to see the solutions to n^n = c .

[@ajaysinghrathore1940]

Well, I was confused because I was expecting a whole number as an answer otherwise I was able to solve this in my mind. I am not good at maths so it was very proud moment for me.

[@user-bo3rz2xd3p]

Математика супер язык! Я не слова не понял о чём говорит автор, но абсолютно всё понял смотря на вывод формул!

[@user-uw7si6pf3w]

Чел... автор неправильно решил задачу потомучто какого хера там получается корень 4-ой степени если х=2 А если хочешь понять че говорит автор, учи инглиш

[@ishanagrawal6399]

I just checked the probable solutions 2^1/2 was smaller 2 was bigger. Remaining was 2^3/4 for clean solution.

[@donmoore7785]

But you missed the infinite number of solutions, and only found one.

[@kobalt4083]

@@donmoore7785 no, only four solutions (x^4=8)

[@gelbkehlchen]

Solution: x^(x^4) = 64 |( )^4 ⟹ x^(4*x^4) = 64^4 ⟹ (x^4)^(x^4) = (8²)^4 = 8^8 |The same number raised to the power of the same number on both sides of the equation ⟹ x^4 = 8 |( )^(1/4) ⟹ x = 8^(1/4) ≈ 1,6818

[@tryfontheofilopoulos1131]

according to the solution x=1,6818 the mistake is 1,6818^1,6818^4=33,O239 NOT 64

[@gelbkehlchen]

@@tryfontheofilopoulos1131 You made a mistake. In this case you must not do (1,6818^1,6818)^4, that is wrong. You must do: 1,6818^(1,6818^4) = 64. Okay?

[@kobalt4083]

Actually, its +/-8^(1/4) since it's an even power.

[@gelbkehlchen]

@@kobalt4083 Yes, you are right!

[@IOwnKazakhstan]

i think they meant to use decimal instead of comma, it's a regional thing@@tryfontheofilopoulos1131

[@gonzalotapia1250]

There is a mistake in 3:45. The property you stated before is absolutely right, but the parenthesis is a must. x^x^4 is absolutely not the same as (x^x)^4. Take 3^3^4 as example. (3^3)^4 = 27^4 = 531.441 3^3^4 = 3^81 = 4.4E38 3^4^3 = 3^64 = 3.4E30

[@MagmaUpwelling]

I can't see where she stated that ( X^X)^4 is equal to (X^X)^4. Do you mean (X^X^4)^4 instead of (X^X)^4?

[@user-di3er2qo3u]

For the equation x^x=64^0.25 x ~= 1.78845 Something somewhere in that area, I'm too lazy to deal with accuracy But this can be done in excel

[@oahuhawaii2141]

x^x^4 is x^(x^4), and not (x^x)^4. That means your analysis is wrong.

[@robertmonroe9728]

Lambert W function to be used

[@KRYPTOS_K5]

I did it without your help. Amazing.

[@user-ly1mn3cp7u]

The following should be noted: The conclusion u^u=8^8 => u=8 is valid because 8>1. The conclusion u^u=a^a => u=a is not valid, if 0

[@PanchalSahib-lh2op]

Legent taking log both side

[@ammuvilambil8032]

x. X xto the power of 4 is 64 that is x to the power of 4 and xto the power of two times put together as x to the power of 2 six times ,that is 64 so x is 2 I may be wrong also

[@user-lf4kg4gg6s]

в прошлом веке когда был маленький в советской школе на олимпиаде решал подобную задачу. (x^x)^4=4. у нас задача красивее.

[@lorenpearson1230]

Wow, that was unnecessarily cumbersome. I took the x^4 part and recognized x had to be a fourth root. Given that I now had (x^(1/4))^((x^(1/4))^4) which simplifies to (x^(1/4))^(x^(1/4 * 4)) = 64. This rewrites as x^(1/4 * x) = 64. I then decomposed 64 to its factor sets and checked them and saw what wold work. 2 fails obviously as too small based upon 2^(1/4 * 2) = sqrt 2. 4 also fails as too small based upon 4^(1/4 * 4) = 4^1 = 4. So I pushed to 8, and it fit as 8^(1/4 * 8) = 8^2 = 64. No extra factor, no pattern matching, just algebra and three test cases. So x is the fourth root of 8 which checks out. If the number was not as factorable as 64 we would have still quickly identified the bounds of x and another tactic would have refined it.

[@forfun4120]

u say that was cumbersome but u literally plugged in values…which in most cases would be much more cumbersome lol

[@donmoore7785]

Your "method" is guessing. lol

[@kobalt4083]

Guessing and checking? Poor.

[@kevincherry4989]

Working this is like going down the rabbit hole of a side trail of a tangent of conversation.

[@honeylin6216]

该题目只要把64变成2的平方的3次方即可。马上就可得出x的4次方等于2的3次方，即等于8.。

[@trysha2340]

Took me about 20 minutes, thinking about various methods. The first once which came to my mind was kinda using fraction, then after not coming to an amswer, I thought about displaying 64 in various ways. then it came to. What is eightnt root of 64 on 4? And by that way, I solved this. You just need to know a little bit of complex roots to finish this.

[@josemauriciomendesdacostam7167]

OBRIGADO PELA EXPLICAÇÃO.!

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@sushant832]

And how do we know that we have to take the power of 4 and not 2 or 8?

[@GabriTell]

I was trying to solve this and I literally accidentally found Euler's number ._.lmao I don't know a lot about advanced maths, so I made a value table of "x^x" and I got to this number making some operations when "x→∞" 👌

[@raushankumar-ic6ve]

at first two times √ both side then x^x=8^1/2 , x=8^1/4

[@sheennina1234]

i just rewrite the exponent x^4=a so my rewritten equation is a^(a/4) = 64 then by inspection 8 satisfies a so x = 8^(1/4)

[@donmoore7785]

But you missed the other solutions...

[@pentagon-math]

Interesting problem, great presentation

[@salahboukerdenna7098]

Note: if power is fractional, we can't take negative solution, but this time yes, x^4 =8 , we can take negative solution.

[@Azif45628]

Since x^4 = x×x×x×x = x×(x^3) x^x^4 = 64 (x^x)^ x^3 = 64 (x^x)^ x^3 = 2^6 (x^x)^ x^3 = (2^2)^3 By equating powers x^3 =3 x= cube root of 3

[@salahboukerdenna7098]

@@Azif45628 How much X⁻½ equals for X=-1/2 ?

[@venomxhawk8427]

@@Azif45628 u messed up from the third line, brackets change the whole meaning

[@Rome3625]

I like how you managed to find correct axe

[@andrasnoll2559]

The root under which the 8 takes place is 4 which is even so the root number can be both - and+ so the+/- was unnecessary to write down. Correct me if I'm wrong.

[@kobalt4083]

Yes. The roots can be both plus and minus. We use +/- to indicate plus or minus. How is it unnecessary?

[@adilbrassem7564]

Ça prouve que les mathématiques est une langue mondiale vivante, que tout le monde, la comprendre facilement.

[@XBGamerX20]

you can use some sort of substitution but the method shown is what id think. idk why I'm watching tho

[@alisterthomas9778]

X b it as 4 quarters tween ❌ b east west north south seen quarter b number 25 or 1/4 b it 4 x or + 25 = 100 however 25 by pronouncing twin 25 as saying 2 x 5 b 10 b the same 64 same b said same quarters of Xs b equal 8s x 4 b 64

[@fishingpepband7786]

So I was dumb and took a bit longer on this but in the process found out a value of x such that x^x^x^x^x^...^x^4=4 which was kind of a weird discovery

[@harji5357]

This problem is so easy I solved it in first sight and i am just an avg scoring indian high schooler

[@robertmonroe9728]

Where is Lambert W-function?

[@Archik4]

You still need to prove that there are no other roots. I think this can be proved through monotonicity.

[@omchavan5664]

Your explanation is good, but i always doubt your methods, i mean it's simple but who can know that if we raise both sides by 4 we will get the answer!!

[@prostatecancergaming9531]

Brilliantly fun problems require brilliantly fun solutions

[@obvioustruth]

And this not one of them.

[@prostatecancergaming9531]

@@obvioustruth elaborate?

[@obvioustruth]

@@prostatecancergaming9531 🤣🤣🤣🤣 I solved x^x^2 = 64 not x^x^4 = 64

[@marceliusmartirosianas6104]

Alternativ metods = X^4= 64 4^16=4^x= x=16 ACADEMIC

[@voyatzo]

You ought to prove that the function x^x is 1-1 for the argument to hold. Cool trick nevertheless.

[@oahuhawaii2141]

There isn't x^x with x = 1, -1 in this problem.

[@kobalt4083]

@@oahuhawaii2141 I think they meant one to one.

[@ManBro25]

How can I use this type of thinking in a daily basis ? What is this ? Abstraction ?

[@alfonbaroify]

It is very complicated. The expresion has the same number X how base and exponet. We can write 64 = 2^ 8 and 8 = 2^ 4 then x^(x^4) = 2^(2^4) then X = 2. That is all. Sorry for my English.

[@ganeshr5]

The numerical answer using log and numerically solving the nonlinear equation is 1.7885. Verify by substituting in the expression x^x^4 = 64.

[@oahuhawaii2141]

But this is wrong. You are not exponentiating properly. x^x^4 is x^(x^4), and not (x^x)^4. The precedence rules for exponentiating is right to left if no parentheses is given to specify the grouping.

[@hectorraulmansilla9064]

Si x⁴ = 8 raíz cuarta de x⁴ = raíz cuarta de 8 lo correcto es | x | = raíz cuarta de 8, el ± está mal

[@soumyraghuwanshi6083]

Taking log both sides will be easier

[@ATTITUDE_KINGS]

(X^4)^x×4=8^8 Then if we take powers then x×4=8 then x=8/4=2

[@pumelo1]

😂😂😂 it is not 2! you are total wrong x^x^4 ----2^2^4....2^16... 2^16 is not 64!!!!! and 2^4 is not 8 but 16🤣🤣🤣Your MAT have ZERO level🤫🤫

[@brownie3454]

2:59 Don’t do math kids. Not even once.

[@alisterthomas9778]

AX B IT THE X SEEING HAVING 4 END POINT Twice therefore b it x4 by x4 by 4 b number solve 4x 4 x 4 b 64

[@JPTaquari]

Dear, I have another way more simple if I didn't make a mistake . I hope, no! X^X^4 = 64 " = (2³)² " = 2^2^3 Bases equal, them x^4 = 2³ X^4 = 8 X = 1,6818

[@oahuhawaii2141]

Wrong. How can you let x = 2 and 4 = 3, and then equate x^4 with 2^3 before simplifying?

[@FlydingVent9296]

I solved it by trial and error with x=2^y/z knowing that x should be a number neither too small or big

[@miantony6493]

kzbin.info/www/bejne/oJ61hZp3drSLo68 You will surely like this math problem

[@Azif45628]

Since x^4 = x×x×x×x = x×(x^3) x^x^4 = 64 (x^x)^ x^3 = 64 (x^x)^ x^3 = 2^6 (x^x)^ x^3 = (2^2)^3 By equating powers x^3 =3 x= cube root of 3

[@kobalt4083]

@@Azif45628 sorry but you made a mistake from x^x^4 = 64 to (x^x)^ x^3 = 64 which led to a wrong solution.

[@AndroidsReview]

I solved like this y=x^4 x=y^1/4 y^y/4=64 y/4*lny=ln64 ylny=4ln64 ylny=24ln2 y=2^z=24/z then brutforce 2 24 4 12 8 8 y=>8 x=> 8^1/4

[@laszlokorosi9012]

How on earth nobody sees that the claimed "solutions" are plain wrong?! Simply check, using the exponential to exponential formula given in the video: (a^m)^n = a^(mn). So (x^x)^4 = x^(4x). Now enter the "solution" 8^(1/4) and you'll get 33.0225..., not 64! The error comes from the third row, where the (x^x^4)^4 expression is erroneously transformed to x^((x^4)4), instead of the correct x^(16x).

[@dominik1204]

it's x^(x^4) not (x^x)^4. Then (2^(3/4))^4 = 2^(4*3/4) = 2^3 = 8 and 2^(3/4)^8 is 2^(8*3/4) = 2^(2*3) = 64

[@THE1-PERCENT-CLUB]

It's simple without formula we can imagine it like 2^2^4=64 , because 2^2=4,now 4^4=64

[@andremenard7124]

if i plug in your answer in excel the result is 33.0225 if the root4 of 8 is equal to 1,681792831 what i am doing wrong

[@LeonardoMenezes03]

Exactly. Something is not right.

[@mihatarataykin]

You are right. The solution on the video is wrong, because (x^4)^(x^4) is not equal to x^4^x^4

[@LeonardoMenezes03]

We have to consider that (x^x)^4 is not the same that x^(x^4). Therefore using x=2^(3/4) in x^(x^4) we get 64. Using in (x^x)^4 we get 33.02252407.

[@andremenard7124]

thank Leonardo for the explanation now i see the form were you get 64

[@MONKEYDLUFFY-vr1fh]

The equation has four roots not two.

[@oahuhawaii2141]

There are 4 easy solutions (2 real, 2 imaginary), and an infinite number of complex solutions. Look up the Lambert function.

[@Feugii]

Or you can juste say : x**x=✓✓64 x**x=2✓2 x=✓(2✓2)