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

@SiwakhileVusimuzi

10 ай бұрын

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

@Taric25

7 ай бұрын

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

@BongiMojapelo

3 ай бұрын

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

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

Жыл бұрын

Unique way of simping Noiceee

@Snoopyguys

Жыл бұрын

Bc

@raid6n529

Жыл бұрын

@@shubhkapoor1940 what is simping?

@miantony6493

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@KSY42

Жыл бұрын

Хороший маркер и всё получится.

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

Жыл бұрын

Can you write the details pls for my little brain ?

@giobur

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

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

Жыл бұрын

@@giobur thanks a lot mate !

@user-gy7zo1eq5z

Жыл бұрын

Yeah it make much more sense

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

Жыл бұрын

Isn't the function f(x) = x^x (i) increasing in x when x > 1 and (ii) satisfying f(x) = 1.

@leweeb949

Жыл бұрын

@@andrewclark2503yulkMK yuko

@FitzTomBlaireau

Жыл бұрын

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

@dudewaldo4

Жыл бұрын

@@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

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

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

@theofigueiredodamasceno5601

Жыл бұрын

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

@musiquinhaslegais4097

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

Жыл бұрын

@@musiquinhaslegais4097 Entendi! Muito obrigado.

@nikitapickf8489

Жыл бұрын

This function is not monotone for x>0

@musiquinhaslegais4097

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

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.

I appreciate the step by step demonstration for a refresher.

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

@mathbook1993

Жыл бұрын

kzread.info/dash/bejne/ZX14wbynoca0etY.html

Daamn That was smooth and clear Well played mate

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

Beautiful handwriting and beautiful explanation!

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

@mathbook1993

Жыл бұрын

kzread.info/dash/bejne/ZX14wbynoca0etY.html

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.

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

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

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

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

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

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

*_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_*

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.

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

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

Жыл бұрын

@@obvioustruth he didn't

@ayushmanchakraborty8838

Жыл бұрын

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

@learf6613

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

Жыл бұрын

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

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

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

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

@user-py6mf9yx2b

Жыл бұрын

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

@MikeN1811

Жыл бұрын

@@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

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.

@user-wz6ec2vo1b

Жыл бұрын

по определению показательной функции основание строго больше нуля

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

OBRIGADO PELA EXPLICAÇÃO.!

@miantony6493

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

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.

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

Nagyon szép volt! Dicsértessék!

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

kzread.info/dash/bejne/ZX14wbynoca0etY.html

The perfection of her hand writing amazed me

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 (which luckily works in this case).

@Commander.and.Chief.Killua

Жыл бұрын

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

@ArloLipof

Жыл бұрын

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

@miantony6493

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@miantony6493

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@carstenmeyer7786

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.*

I like how you managed to find correct axe

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

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? ;-)

Nice problem and good explanation!

@obvioustruth

Жыл бұрын

"good explanation" for dumb losers e to make them dumber!!! 🤣🤣🤣🤣

Outstanding and excellent

Impressive ❤️🔥

Interesting problem, great presentation

Brilliantly explained

@miantony6493

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

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

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

kzread.info/dash/bejne/ZX14wbynoca0etY.html

@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.

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.

Thank you very much

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

kzread.info/dash/bejne/ZX14wbynoca0etY.html

@froreyfire

10 ай бұрын

Artistic. :-)

@froreyfire

10 ай бұрын

Artistic. :-)

@varoonnone7159

10 ай бұрын

​@@froreyfire Maths doesn't have a god

@froreyfire

10 ай 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".

THIS IS A GOOD ONE !!!

😊👍👍👍Very good thank you

I did it without your help. Amazing.

Great explanation

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.

Woww thanks very good tyy

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.

Excellent

Brilliantly fun problems require brilliantly fun solutions

@obvioustruth

Жыл бұрын

And this not one of them.

@prostatecancergaming9531

Жыл бұрын

@@obvioustruth elaborate?

@obvioustruth

Жыл бұрын

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

Very instructive task

very nice question and a great solution

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)

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

x = 2^(3/4)

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

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)

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

Thanks

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.

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

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@Azif45628

Жыл бұрын

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

Жыл бұрын

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

Did my usual iterative "method"..... very tedious and not a patch on your method. Neat one😀

@sapri344

Жыл бұрын

what do you mean by "iterative method"

@Azif45628

Жыл бұрын

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

@oahuhawaii2141

Жыл бұрын

@Rehaan: Why are you posting the same wrong "solution" all over the Comments section? You should just post one, and ask someone to find where you messed up. And delete all those other copies.

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

@Musabll78

Жыл бұрын

W^M Faster!

can we solve it by taking log base8 on both side

@miantony6493

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@oahuhawaii2141

Жыл бұрын

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

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

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@user-im3pf1fs4p

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

@user-mh6ke4zs4c

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.

I like the channel idea

Wonderful

Lambert W function to be used

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

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

@user-uw7si6pf3w

11 ай 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

Спасибо.

Legent taking log both side

I have noooooo idea what I just watched, but I'm glad I did! Thank you so much! Or is just that I'm too high 🤔? Either way great video!

@cryptowalls

Жыл бұрын

did you catch the explanation @4:20 ... ... ... ;)

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

Good trick

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

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→∞" 👌

I really like the way you write an X

Must convert to decimal Good job

Nice...

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 .

Nice logic

Very good

@obvioustruth

Жыл бұрын

Very good idiot!

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

Taking log both sides will be easier

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.

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

very nice

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.。

The equation has four roots not two.

@oahuhawaii2141

Жыл бұрын

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

Muito bom

Genius

Axe to the power axe to the power 4. Good question. Axe is important.

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...

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!!

Nice video

Love math❤️

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

Жыл бұрын

kzread.info/dash/bejne/nKGGtsl6cbK-nbA.html You will surely like this math problem

@Azif45628

Жыл бұрын

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

Жыл бұрын

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

@Azif45628

Жыл бұрын

@@JiminatorPV in such case powers couldn't equate

@JiminatorPV

Жыл бұрын

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

2^2^4

Nice