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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 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!
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
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.
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.
@@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
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
@@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
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.
*_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_*
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.
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.
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).
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.
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 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
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.
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? ;-)
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).
@@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.*
@@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.
@@MikeN1811 Подставьте -8^0.25 в условие. Получите тождество. А то, что не стыкуется с промежуточными выкладками, проблема этих выкладок, а не результата.
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.
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.
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.
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
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.
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).
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.
@@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.
@@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.
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 .
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.
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 ❤
@@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".
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
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..)
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
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
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.
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.
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
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→∞" 👌
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
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.
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?????????
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
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.
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!!
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
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 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?
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.
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
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
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
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?
Как увидел задачку на превью сразу открыл калькулятор и начал возводить 8 в степень 1/4, то есть, 8^0.25 как то интуитивно понял что "х" равен этому значению.
