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

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

[mathwindow]

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

[spymadmax]

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

[hydraslair]

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

[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.

[artkirakosyan]

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.

[musiquinhaslegais]

One little detail: it is useful to note that the function (x^4)^(x^4) is monotone for x>1 and is <1 for |x|<1 (then the only positive solution is the one you found, analogously there is only one negative solution too, so the only real solutions are the ones you found). You showed us a very clever trick here! I am enjoying your videos a lot!
(Edit: not monotone for x>0, but certainly is for x>1)

[kobalt]

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).

[timeonly]

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? ;-)

[errornotfound_]

*_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_*

[ejehzyq]

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

[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

[MikeN]

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

[michaeledwards]

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.

[jim]

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

[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.

[phuocvlog]

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

[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).

[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.

[b-penajohneric]

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