If you have any suggestions or questions on math, comment as a reply! ❤😊
@SiwakhileVusimuzi
Жыл бұрын
Am having difficulties with this x^x^27=77 😢
@Taric25
11 ай бұрын
You forgot the imaginary solutions ±(⁴√8)i.
@BongiMojapelo
7 ай бұрын
So why did the person say( x^ x4)4 where did the get the 4 in the first solving stage
@spymadmax584
2 жыл бұрын
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
2 жыл бұрын
Unique way of simping Noiceee
@Snoopyguys
2 жыл бұрын
Bc
@raid6n529
2 жыл бұрын
@@shubhkapoor1940 what is simping?
@miantony6493
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@KSY42
2 жыл бұрын
Хороший маркер и всё получится.
@hydraslair4723
2 жыл бұрын
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
2 жыл бұрын
Can you write the details pls for my little brain ?
@giobur
2 жыл бұрын
@@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
2 жыл бұрын
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
2 жыл бұрын
@@giobur thanks a lot mate !
@荣杰陈
2 жыл бұрын
Yeah it make much more sense
@bluefkingstar
2 жыл бұрын
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
2 жыл бұрын
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
2 жыл бұрын
@@andrewclark2503yulkMK yuko
@FitzTomBlaireau
2 жыл бұрын
@@bluefkingstar Thanks for supplying this insight, it's much more satisfying than this vid (which I clicked on expecting a method)
@dudewaldo4
Жыл бұрын
@@bluefkingstar 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
Жыл бұрын
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
2 жыл бұрын
One little detail: it is useful to note that the function (x^4)^(x^4) is monotone for x>1 and is 1)
@theofigueiredodamasceno5601
2 жыл бұрын
Não entendi o que você quis dizer com monótono pra x>0, explica pfpf.
@musiquinhaslegais4097
2 жыл бұрын
@@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
2 жыл бұрын
@@musiquinhaslegais4097 Entendi! Muito obrigado.
@nikitapickf8489
2 жыл бұрын
This function is not monotone for x>0
@musiquinhaslegais4097
2 жыл бұрын
@@nikitapickf8489 I think you're right. Correcting it: if |x|=1, but since the function iis monotone for x>1 and for x
@artkirakosyan2633
2 жыл бұрын
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
Жыл бұрын
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.
@errornotfound_1972
Жыл бұрын
*_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_*
@Zwerggoldhamster
2 жыл бұрын
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
2 жыл бұрын
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
2 жыл бұрын
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
2 жыл бұрын
@@jyotismoykalita 0^0 is not "undefined", it is 1. x^0 is equal 1 for any x you choose, positive, negative or zero.
@riggsmarkham922
2 жыл бұрын
@@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
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@jim2376
Жыл бұрын
The precision and readability of her writing are amazing. And multicolored for a bonus!
@mathbook1993
Жыл бұрын
kzitem.info/news/bejne/lH96xY6kqpV_fag
@OrenLikes
Жыл бұрын
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.
@VeteranVandal
2 жыл бұрын
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
2 жыл бұрын
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
2 жыл бұрын
@@obvioustruth he didn't
@ayushmanchakraborty8838
2 жыл бұрын
@@obvioustruth it's not wrong, go to junior school and learn about exponents, bloody illiterates
@learf6613
2 жыл бұрын
@@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
2 жыл бұрын
I was trying to guess at it too, and eventually ended up at 4th root of 8, which is same as you.
@ArloLipof
2 жыл бұрын
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
2 жыл бұрын
Try (+or-square root of 2square root of 2)^8
@ArloLipof
2 жыл бұрын
@@Commander.and.Chief.Killua Please only comment if you understand the needed concepts of mathematics. What do you mean?
@miantony6493
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@miantony6493
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@carstenmeyer7786
2 жыл бұрын
@@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.*
@michaeledwards2251
2 жыл бұрын
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
Жыл бұрын
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.
@Balila_balbal_loki
2 жыл бұрын
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.
@phuocvlog
2 жыл бұрын
we just imply x^4=8 from (x^4)^(x^4) =8^8 when function y=x^x is monotone
@duke6585
Жыл бұрын
As a 7th grader, I can say with confidence I’m ready for algebra 7836. Thanks for the quick lessons!
@mangouschase
Жыл бұрын
i'm sorry but you are not, this things are just 10th grade (if i did the conversion correctly to american)
@OSU23901
Жыл бұрын
@@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
Жыл бұрын
lmao you go kid
@comptech5240
Жыл бұрын
@@mangouschase in all seriousness, someone who has learned exponents nicely in 7th grade will be able to do it easily in 8th grade
@MikeN1811
2 жыл бұрын
Уравнение имеет только один положительный корень 8^(1/4), так как второй отрицательный и не входит в область определения функции.
@Анна-п4й4ж
Жыл бұрын
А почему не входит? Даже если икс отрицательный, то парная степень минус убирает
@MikeN1811
Жыл бұрын
@@Анна-п4й4ж 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
Жыл бұрын
@@MikeN1811 Подставьте -8^0.25 в условие. Получите тождество. А то, что не стыкуется с промежуточными выкладками, проблема этих выкладок, а не результата.
@ОлегКонев-ъ2ь
Жыл бұрын
по определению показательной функции основание строго больше нуля
@b-penajohneric151
2 жыл бұрын
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
Жыл бұрын
kzitem.info/news/bejne/lH96xY6kqpV_fag
@ΓιωργοςΡακοπουλος-δ2ο
Жыл бұрын
The 8^(1/4) is ~~ 1,681739, which X^X^4 results in ~~ 33,02252 and not 64. What i am doing wrong?
@oahuhawaii2141
Жыл бұрын
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.
@bmbelko
2 жыл бұрын
I appreciate the step by step demonstration for a refresher.
@GDyoutube2022
Жыл бұрын
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
Жыл бұрын
kzitem.info/news/bejne/lH96xY6kqpV_fag
@oahuhawaii2141
Жыл бұрын
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
Жыл бұрын
@@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
Жыл бұрын
@@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.
@tgx3529
Жыл бұрын
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)
@victorguilherme7955
Жыл бұрын
I just found your channel right now and loved it. Very helpful! Thanks from Brazil!
@mathbook1993
Жыл бұрын
kzitem.info/news/bejne/lH96xY6kqpV_fag
@sangjeonglee4182
8 ай бұрын
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
@Anonymous.s.r
2 жыл бұрын
can we solve it by taking log base8 on both side
@miantony6493
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@oahuhawaii2141
Жыл бұрын
Well, you didn't try it because that doesn't work out.
@RileyRampant
Жыл бұрын
your handwriting is beautiful. Its a joy just to see you write these expressions.
@serhiislobodianiuk776
2 жыл бұрын
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
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@@Azif45628 you asumed that x=2 by equating exponents
@Azif45628
2 жыл бұрын
@@JiminatorPV in such case powers couldn't equate
@JiminatorPV
2 жыл бұрын
@@Azif45628 that's what i mean, you can't equate powers here
@oahuhawaii2141
Жыл бұрын
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 .
@timeonly1401
6 ай бұрын
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? ;-)
@joshmckinney6034
2 жыл бұрын
Beautiful handwriting and beautiful explanation!
@strayfox6835
Жыл бұрын
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
@shlokee
Жыл бұрын
if we use differentiation (or operator and log) it'll be a lil different
@arjunsanap378
Жыл бұрын
Take log will be easy I think so
@tmp3477
2 жыл бұрын
So, axing the exponent of your ex, gives your eggs in hex. Very clear.
@antilex07
2 жыл бұрын
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
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@ВоваГуд-ъ5и
Жыл бұрын
точно, влепить в уравнение не 64, а 65 или 63, накувыркалась бы с корнями различных степеней, это не железный алгоритм, из носа выковырянный
@ЕремейЗонов
Жыл бұрын
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.
@altSt0rm
2 жыл бұрын
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
Жыл бұрын
NO если в задаче не указан тип переменной то он считается вещественным и не может иметь комплексное значение
@lukandrate9866
Жыл бұрын
@@eliremingtone4654 "if you consider the complex plane as well" - не, не видели
@eliremingtone4654
Жыл бұрын
@@lukandrate9866Does the problem mention the complex plane? Есть простое правило -- если ничего не указано числа считаются вещественными
@oahuhawaii2141
Жыл бұрын
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
Жыл бұрын
For x^4=8, the only complex solutions are x=+/-4th root of 8 * i.
@gonzalotapia1250
Жыл бұрын
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
Жыл бұрын
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?
@hieuduong7447
Жыл бұрын
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
Жыл бұрын
kzitem.info/news/bejne/lH96xY6kqpV_fag
@froreyfire
Жыл бұрын
Artistic. :-)
@froreyfire
Жыл бұрын
Artistic. :-)
@varoonnone7159
Жыл бұрын
@@froreyfire Maths doesn't have a god
@froreyfire
Жыл бұрын
@@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".
@str8l1ne44
2 жыл бұрын
Daamn That was smooth and clear Well played mate
@darkknight2414
2 жыл бұрын
I dont know why I get these videos recommended at 3am and it leads me to try to solve this at night🙄
@alittax
Жыл бұрын
Thank you, that was very satisfying to watch! Nice work!
@trysha2340
Жыл бұрын
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.
@lorenpearson1230
2 жыл бұрын
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
2 жыл бұрын
u say that was cumbersome but u literally plugged in values…which in most cases would be much more cumbersome lol
@donmoore7785
Жыл бұрын
Your "method" is guessing. lol
@kobalt4083
Жыл бұрын
Guessing and checking? Poor.
@tanvirahmed1033
Жыл бұрын
Outstanding and excellent
@КлимСамгин-р6х
2 жыл бұрын
в прошлом веке когда был маленький в советской школе на олимпиаде решал подобную задачу. (x^x)^4=4. у нас задача красивее.
@ManBro25
10 ай бұрын
How can I use this type of thinking in a daily basis ? What is this ? Abstraction ?
@honeylin6216
Жыл бұрын
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.。
@sushant832
2 жыл бұрын
And how do we know that we have to take the power of 4 and not 2 or 8?
@ishanagrawal6399
2 жыл бұрын
I just checked the probable solutions 2^1/2 was smaller 2 was bigger. Remaining was 2^3/4 for clean solution.
@donmoore7785
Жыл бұрын
But you missed the infinite number of solutions, and only found one.
@kobalt4083
Жыл бұрын
@@donmoore7785 no, only four solutions (x^4=8)
@panosdiamadopoulos1682
2 жыл бұрын
W lambert function and is solvable in 5 lines without literally any thinking
@Musabll78
2 жыл бұрын
W^M Faster!
@hedwig7526
Жыл бұрын
The perfection of her hand writing amazed me
@pentagon-math
Жыл бұрын
Interesting problem, great presentation
@ajaysinghrathore1940
2 жыл бұрын
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.
@НиколайГолубцов-ч6и
Жыл бұрын
Математика супер язык! Я не слова не понял о чём говорит автор, но абсолютно всё понял смотря на вывод формул!
@ИванБорисов159
Жыл бұрын
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
@andrasnoll2559
Жыл бұрын
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
Жыл бұрын
Yes. The roots can be both plus and minus. We use +/- to indicate plus or minus. How is it unnecessary?
@XBGamerX20
2 жыл бұрын
you can use some sort of substitution but the method shown is what id think. idk why I'm watching tho
@salahboukerdenna7098
2 жыл бұрын
Note: if power is fractional, we can't take negative solution, but this time yes, x^4 =8 , we can take negative solution.
@@Azif45628 u messed up from the third line, brackets change the whole meaning
@syberrus
2 жыл бұрын
-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
Жыл бұрын
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)) .
@СлободанДрамлић
Жыл бұрын
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
@GabriTell
Жыл бұрын
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→∞" 👌
@Luffy_wastaken
2 жыл бұрын
So if some asks me, 3^3^3 Do I interpret it as 3^9 or 3^27?
@hanskywalker1246
2 жыл бұрын
I think only when its in brackets, so (3^3)^3=3^3*3=3^9, if its 3^3^3 then 3^27
@amongus2816
2 жыл бұрын
It depends on where the parentheses go, if it looks like (3^3)^3 then you would use the power rule to get 3^9 but if it was 3^(3^3) then it would be 3^27
@iarmycombo5659
2 жыл бұрын
Either 27^3 or 3^9, but never 3^27 because the exponents are always multiplied. Edit: Only ever is it 3^27 if the original version is 3^(3^3) but never if its 3^3^3
@Simio_Da_Tundra
2 жыл бұрын
@@iarmycombo5659 actually no, if you have no parenthesis, the order of tetration is from the top exponent down, so, with no parenthesis, 3^3^3=3^27
@iarmycombo5659
2 жыл бұрын
@@Simio_Da_Tundra wow, i guess i just translated it straight to 3² where 2 = 3³
@kusalachandrabehera4255
Жыл бұрын
Thank you very much
@sourivore
2 жыл бұрын
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
Жыл бұрын
Lambert function shows that there are an infinite number of complex solutions.
@gelbkehlchen
Жыл бұрын
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
Жыл бұрын
according to the solution x=1,6818 the mistake is 1,6818^1,6818^4=33,O239 NOT 64
@gelbkehlchen
Жыл бұрын
@@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
Жыл бұрын
Actually, its +/-8^(1/4) since it's an even power.
@gelbkehlchen
Жыл бұрын
@@kobalt4083 Yes, you are right!
@IOwnKazakhstan
Жыл бұрын
i think they meant to use decimal instead of comma, it's a regional thing@@tryfontheofilopoulos1131
@josemauriciomendesdacostam7167
2 жыл бұрын
OBRIGADO PELA EXPLICAÇÃO.!
@miantony6493
2 жыл бұрын
kzitem.info/news/bejne/y6OIupt3eoGJoII You will surely like this math problem
@ВасяПупкин-б1м9з
2 жыл бұрын
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
Жыл бұрын
x^x^4 is x^(x^4), and not (x^x)^4. That means your analysis is wrong.
@yungifez
2 жыл бұрын
Are calculators allowed? If yes, take the log of both sides
@oahuhawaii2141
Жыл бұрын
Won't help. Try it and you won't get to the solution any faster.
@sheennina1234
Жыл бұрын
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
Жыл бұрын
But you missed the other solutions...
@linksmath124
Жыл бұрын
very nice question and a great solution
@vlsext
Жыл бұрын
what about x^(x^3.89) ? what about x^(x^y)? In assembly language, please. Or, at least, in C/C++ which is also good for microcontrollers. Oh, you have additional question? About precision? Or about performance?
@philippeb.8677
Ай бұрын
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
@voyatzo
2 жыл бұрын
You ought to prove that the function x^x is 1-1 for the argument to hold. Cool trick nevertheless.
@oahuhawaii2141
Жыл бұрын
There isn't x^x with x = 1, -1 in this problem.
@kobalt4083
Жыл бұрын
@@oahuhawaii2141 I think they meant one to one.
@omchavan5664
Жыл бұрын
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!!
@ammuvilambil8032
Жыл бұрын
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
@ganeshr5
2 жыл бұрын
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
Жыл бұрын
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.
@alisterthomas9778
Жыл бұрын
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
@robertmonroe9728
Жыл бұрын
Lambert W function to be used
@NTC_whitecrayon
2 жыл бұрын
Working this is like going down the rabbit hole of a side trail of a tangent of conversation.
@meerayft824
2 жыл бұрын
how about we start with taking log on both sides and simplifying....... ofc ideas are plenty but implementation? nah...... idk.. tbh im lazy so yeah... can we solve using logarithms?????????
@jeroensoenen4054
2 жыл бұрын
Am I wrong or are +/- i 8^(1/4) also solutions?
@ttrgan
2 жыл бұрын
no, you're right
@sallesvianagomesdemagalhae6181
2 жыл бұрын
You are imagining things
@KRYPTOS_K5
Жыл бұрын
I did it without your help. Amazing.
@jmforet27
Жыл бұрын
Hi, I did the same, but at the end 8^1/4 is close to 1.6818. If I put 1.6818 it into the equation, it gives 33 .xxx not 64. The solution should be close to 1.78xxx . I don't see where is the error ....
@billr3053
Жыл бұрын
The rule for such chained exponents is that you do the uppermost first and work your way down. So that’s 1.68179 to the exponent 4 equals 8. And then 1.168179 to the exponent 8 = 64.
@fishingpepband7786
2 жыл бұрын
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
@math_qz_2
9 ай бұрын
Very instructive task
@robertmonroe9728
Жыл бұрын
Where is Lambert W-function?
@DrLiangMath
2 жыл бұрын
Nice problem and good explanation!
@obvioustruth
2 жыл бұрын
"good explanation" for dumb losers e to make them dumber!!! 🤣🤣🤣🤣
@raushankumar-ic6ve
Жыл бұрын
at first two times √ both side then x^x=8^1/2 , x=8^1/4
@TheColoursofMathematics
Жыл бұрын
Thanks
@ATTITUDE_KINGS
Жыл бұрын
(X^4)^x×4=8^8 Then if we take powers then x×4=8 then x=8/4=2
@pumelo1
Жыл бұрын
😂😂😂 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🤫🤫
@Musabll78
2 жыл бұрын
CommodoreC64=x=11=3=32*2=64
@alfonbaroify
Жыл бұрын
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.
@violeton
2 жыл бұрын
Is process of elimination a thing?
@qazwsxqaz3163
Жыл бұрын
Где это можно применить в жизни??? Where in real life to apply this example?
@wedey4you360
Жыл бұрын
Hello 👋 madam can I know the font type you use for your thumbnails
@Rome3625
Жыл бұрын
I like how you managed to find correct axe
@antoniusnies-komponistpian2172
Жыл бұрын
Though I'm not convinced yet that a=b is the only solution for a^a=b^b
@oahuhawaii2141
Жыл бұрын
Look up the Lambert function to see the solutions to n^n = c .
@PanchalSahib-lh2op
10 ай бұрын
Legent taking log both side
@kindrom
Жыл бұрын
This is a Math Olympia question, and of high school? I am pretty sure ASME and AIME and US Olympiad tests are way harder than those, even in the 90s when I took them.
@ΓιωργοςΡακοπουλος-δ2ο
Жыл бұрын
if we work little diferent { xln(4)=ln(64) and more} we find that x~~1,78845416 which fits everywhere.
Пікірлер: 825