There are 3 solutions. You got just one because you are assuming x^(1/x) as an injective function. Its not. The other 2 real solutions can be expressed with W-Lambert function.
Tell me one reason, I shouldn't use logarithms' properties to solve this equation, rather than using my intuitive brain to find out a creative look on a simple problem and maje it much more complex than it looks.
Yes, and one between -1 and 0. To find them you have to use the Lambert W function. 2^x = x^32 (e^ln(2))^x = x^32 (write 2 to base e) e^(x*(1/32)*(ln(2)) = x^1 (raise both to 1/32 power) e^(-x*(1/32)*(ln(2)) = x^-1 (raise to negative 1 power) xe^(-x*(1/32)*(ln(2)) = 1 (multiply both by x) -x*(1/32)*ln(2) * e^(-x*(1/32)*(ln(2)) = -ln(2)/32 (multiply by -ln(2)/32 to have the exponent and scalar be the same) -x*(1/32)*ln(2) = W(-ln(2)*(1/32)) (Apply Lambert W function) x = -32/ln(2) * W(-ln(2)/32) Depending on which branch of the Lambert W function you apply you will get the two missing solutions which are Approx x = 1.0224 Approx x = -0.979
SON ELOCUENTES LAS CRITICAS AHORA YO LES PREGUNTO QUE SIGNIFICA QUE HAYA UNA FORMA DE SOLUCIONARLA ASI QUISAS FALTA UNA RESTRICCION AL INICIO SE OLVIDO PONERLO AY SI VALDRIA CREO QUE FUE ESO POR ESO SI ESTARIA BIEN
It's not possible because it's a transcendental equation where x is in the exponent and on the other hand written as an argument and that at the same time. In generall, this type of equation can be solved numerically or in some cases with the W Lambert function.
this is the difference between a proof and a problem solving technique, a proof looks elegant and persuasive but if you had the same problem yourself you wouldn't be able to solve it.
Somehow this video got a lot of criticism. I enjoyed it and learned something. OK, there is another solution, but what is great about this video is that you don't have to roll out the big guns of the Lambert W function.
There are 3 real solutions. He just got only one because he's assuming x^(1/x) as an injective function. It's not. The other 2 solutions can be found with lambert function.
Would not work, you would need to use the product logarithm, which is a function specifically designed to tackle problems like xln(x), which would occur if you try to solve this problem using a logarithm
3 solutions: 256, and two more between -1 and 2. All you have to do is to substitute x = -1, 0, 2 and see that functions x^32 are low in the middle and high on the sides.
this trick can only be aplied for 2^x= x^(2^p) with p being any of the pell numbers. a similar trick can be used for any base [ y^x=x^(y^z) ] but I didn't bother to check which sequences produce z for it to be true. that being said, I expect those sequences to be similar to the golden ratio or the pell numbers.
If you use the LambertW-function you will get around 1.02239 (I'm very exact since it concerns such a small number being impacted by 2 extremely different expressions).
The W-function is a type of logarithm that works for x×e^x. You know that ln(e^x)=x, the responding way of only getting x from x*e^x is writing W(x*e^x)=x. As long as both the factor and the exponent are identical you can use W successfully.
Even if you have (x/2)*e^(x/2) in an eqaution, you will get the factor in front of the exponential expression (being e^(x/2)). Thus, W((x/2)*e^(x/2))=x/2. With the fundementals understood, we can solve the equation in the video.
Basically, your goal in using the W-function is to get n*e^n=I, with n being any expression involving x and I being an integer. 2^x=x^32 (2^x)^1/32=(x^32)^1/32 2^(x/32)=x 1=x/(2^(x/32)) 1=x*2^(-x/32) 1=x*(e^ln(2))^(-x/32) 1=x*e^(-(ln(2)*x/32)) -ln(2)/32=-ln(2)*x/32*e^(-ln(2)*x/32) W(-ln(2)/32)=W(-ln(2)*x/32*e^(-ln(2)*x/32)) W(-ln(2)/32)=-ln(2)*x/32 W(-ln(2)/32)*32/(-ln(2))=x Put this expression in Wolframalpha, instead of W, you write LambertW(numbers).
vabb ma non è un metodo.. che ne sai quando trovi la combinazione giusta? potrebbe essere dopo 2 tentativi come dopo 1 miliardo. Si passa al log in base 2, poi applicando la formula del cambio di base si semplifica.
Logarithms: ok ok I’m useless
IKR!
So trial and error in the end.
😢😢
How would you use logs to do it?
True
for all people suggesting logs
you try to solve it with logs, you can't.
you need logs and lambert W function
@@lolok6439 right
There are 3 solutions not just 256
Also 1.02239 and -0.979017
@@FuturePast2019 These are approximate, not true solutions
@@DEXyk São soluções pois encostam no eixo X.
We should also prove there is no other solutions existing.
There are 2 more
that's asking too much of people like this.
Why multiply with 1/32X..
I mean that's alright, but why the X .???
Please explain..
Because he wants to create an equation with x values one side and numerical values with another side
He just forgot to prove 0 is not a solution, because if it was, it would have been impossible
There are 3 solutions. You got just one because you are assuming x^(1/x) as an injective function. Its not. The other 2 real solutions can be expressed with W-Lambert function.
awesome
Blackpenredpen enjoyer?
@@breadles5 dont know what that is
@canalf007 probably one of the most fun math teachers on UA-cam, i learned about the Lambert W function from him.
@@breadles5 oh ok, i'll check it out
Tell me one reason, I shouldn't use logarithms' properties to solve this equation, rather than using my intuitive brain to find out a creative look on a simple problem and maje it much more complex than it looks.
Lambert w and logs
Great
少なくとも負にもう一個解あるよね
Mantap❤❤❤
What if 2^x = x^31 ?
Shouldn't there be a value of x between 1 and 2?
Yes, and one between -1 and 0. To find them you have to use the Lambert W function.
2^x = x^32
(e^ln(2))^x = x^32 (write 2 to base e)
e^(x*(1/32)*(ln(2)) = x^1 (raise both to 1/32 power)
e^(-x*(1/32)*(ln(2)) = x^-1 (raise to negative 1 power)
xe^(-x*(1/32)*(ln(2)) = 1 (multiply both by x)
-x*(1/32)*ln(2) * e^(-x*(1/32)*(ln(2)) = -ln(2)/32 (multiply by -ln(2)/32 to have the exponent and scalar be the same)
-x*(1/32)*ln(2) = W(-ln(2)*(1/32)) (Apply Lambert W function)
x = -32/ln(2) * W(-ln(2)/32)
Depending on which branch of the Lambert W function you apply you will get the two missing solutions which are
Approx x = 1.0224
Approx x = -0.979
@@XtronePlaysG Yes there is one solution below 2. It can be obtained with other methods.
@@henrikstenlund5385 what other method could you use other than the Lambert W function?
@@XtronePlaysG I used differentiation
@@henrikstenlund5385 differentiating both sides doesn't maintain equality though
also 1.02239 and -0.979017
1.02239 it's not a correct answer. If you have calculator check it.
@@hrithikroshanfan4664 it is a correct answer
Onde estão as outras 31 soluções bro...😶🌫️
Trial-and-error. I'd rather not calling it a math. It makes me look bad. And you?
何も役に立たない、必要十分条件を学び直してください。
Polarising
Yep, I understand why I only got a B in Calc now 😅
This is just a simple exponential equation. It is not even Calculus.
視覚的解析法に持っていくんですね
SON ELOCUENTES LAS CRITICAS AHORA YO LES PREGUNTO QUE SIGNIFICA QUE HAYA UNA FORMA DE SOLUCIONARLA ASI QUISAS FALTA UNA RESTRICCION AL INICIO SE OLVIDO PONERLO AY SI VALDRIA CREO QUE FUE ESO POR ESO SI ESTARIA BIEN
please solve this one 2^x=x^8
1. Assume x > 0
2. Replace x with 2^y
3. 2^(2^y)=(2^y)^32 2^(2^y)=2^(32y) 2^y=32y
4. Real solution for y without W function is 8, so x=2^8=256
Y^32 means
(y X y X y …) 32 times
It is not adding so how can you say 32y
It would be called 32y if
(y+y+y…) 32 times
So your steps are wrong.
Solve, 3^x=x^9
Find x?
Why 2/64 we have?!
X product both sides 😵💫😵💫😵💫😶😶😶😶😶😶😶😶😶
konchem clear ga explain cheyyandi
I got lost 😢
😮
Just use logarithms.
Yea but how?
It's not possible because it's a transcendental equation where x is in the exponent and on the other hand written as an argument and that at the same time.
In generall, this type of equation can be solved numerically or in some cases with the W Lambert function.
Não entendi 2/64?
Very easy 256
very incomplete too
@@snooker216 🤣🤣🤣🤣
X=256
I don't think this is correct. I think the answer should be 2 and 1/16
damn, i went 5, 32, ohhhhhh
Bro what?
응?
너무 무식한 방법인데?
this is the difference between a proof and a problem solving technique, a proof looks elegant and persuasive but if you had the same problem yourself you wouldn't be able to solve it.
I don't like this video. Seems like you already have to know the answer to solve it this way.
Он просто показал, как уравнять основание и степень. И приравнять х^х.
Probably this way is for those who had never been taught logarithm;otherwise there is No reason why we use this unrealistic process.
There's another answer if you draw their graphs. It's in (-1,0) although I don't know the exact value
@@elgb5671 its aprox. -0.979 i think, you have to use lambert W function for that
log : im a joke for you
xを整数と仮定するならこの式を満たすxは1つ存在し256である。
(証明)
x≦0のとき、2ˣは単調増加、x³²は単調減少であり、x=-1のとき、2ˣx³²であるから、求めるxはx>0である。
この時、両辺に自然対数をとって変形すると
(logx)/x=(log2)/2⁵
右辺の分母分子にn(n∈ℤ)を掛けて両辺を比較する。x=2ⁿ=n･2⁵となればよい。
ここで、n=2ᵏ(k∈ℤ)と表すことができるから、2ᵏ=k+5を満たすkを求めればよい。
この式を満たすkはk=3
よって、n=2³=8であるから
x=2⁸=256
(証明終)
wa ta shi mo >>> n=2³=8であるから
x=2⁸=256
大学受験レベルだと
2^k=k+5が1つの解しか持たないことの証明が必要(自明ではあるが)
これなら
k=256のときこの式が成り立つ■
で証明が完了してしまう
指数って表示できるんだ初めて知ったｗ
Can you please explain this further@@ちょもらんま-o8p
Chow ni. Chow chow???
Not clearly understand
Please see my video for more explanation.
ua-cam.com/video/4e-SV9sRTt4/v-deo.htmlsi=gTAMd1T4ij2O6ogo
logarithms would have eliminated the trial and error
@@PixalonGC how to use log in this question
3 real solutions. You only got one of them.
How do you know there are 3 real solutions
@@noname-ed2un x≈1.02239, x≈-0.979017, x=256. Graph it and it is as clear as day. Values can be found leveraging the lambert W function.
2 real solutions, 1.022 and 256
@@spuckhafte And a third, x=-0.979017
Às syuçh no high theorem exist like this 😊
Wow I did not expect that you would solve it that way. I would use logarithms on that 😮
Somehow this video got a lot of criticism. I enjoyed it and learned something. OK, there is another solution, but what is great about this video is that you don't have to roll out the big guns of the Lambert W function.
Gotta use Lambert W to solve this.
It is wrong answer, also there is another x, you should use W function of lambert.
全く役に立たない、誤解を与える解法ですね。
解が一つ見つかったけど、これ以外に解がないと言うのは明らかか？
他に実数解があるかは分からないですけど、この解法だと解が一つだと示せてないので✖︎ですね
f(x):=2^x-x^32
f(1)=2-1>0
f(2)=2^2-2^32
There are 3 real solutions. He just got only one because he's assuming x^(1/x) as an injective function. It's not. The other 2 solutions can be found with lambert function.
x가 음수인 해는 고려하지 않는 풀이법이네요
Graph says there's another solution
This happens when just play songs and don’t explain how it’s done
Detailed solution video here.
ua-cam.com/video/4e-SV9sRTt4/v-deo.htmlsi=gTAMd1T4ij2O6ogo
これをなぜ堂々と載せられるんだ
Why
Solo obtuviste la solución trivial, Faltan las otras soluciones, que pasó??
It would be more easy to find the value by applying log It seems to me
Please explain
Would not work, you would need to use the product logarithm, which is a function specifically designed to tackle problems like xln(x), which would occur if you try to solve this problem using a logarithm
最後がなぜ成立するの？十分性は？
Without excluding x non zero , that is wrong procedure. Use log.
Music is quite boring my dear,this is classroom, not a cinema, please remove the music 😊
U gotta put a cause x ≠0
זה לא הוגן . מאיפוא צץ לו 1/32x ?
3 solutions: 256, and two more between -1 and 2. All you have to do is to substitute x = -1, 0, 2 and see that functions x^32 are low in the middle and high on the sides.
@@megavitek , try to verify with x = {-1, 0, 2}.🤔
x1=1.0224
x2=-0.97902
x3= 256
Me : 2^0 = 1^32 ☠️
is this satire
not that nice, clearly unpredictable
And 32 more solution
How 2 power 1/32 becomes 2/64
You multiply the power by 2/2 which is the same as multiplying by 1, its an equivalent fraction
Because 2/64 = 1/32
Facil demais x=256
it is not logically accepted
256 by trail and error
Не правильно
Logを使おう
I can't see your letters
148
Not good. It’s not math.
NÃO ENTENDI
It's correct when X#0
Je ne comprends rien
X=32 is also a root
x= 32 sir
São 3 soluções.
x=256
x = - 32 LambertW(- ln(2)/32)/ln(2)
x = - 32 LambertW(ln(2)/32)/ln(2)
Why immediately jump from (16)^(2/256) to (256)^(1/256)? Makes no sense.
Because (16)^2 is 256.
@@joaquimeiras3367 At what point did they explain that was their logic?
В России это называется "метод научного тыка". Было интересно? Ставь лайк
Еще должен быть один отрицательный корень уравнения
7:27 Veterinary Surgeon
this trick can only be aplied for 2^x= x^(2^p) with p being any of the pell numbers.
a similar trick can be used for any base [ y^x=x^(y^z) ] but I didn't bother to check which sequences produce z for it to be true.
that being said, I expect those sequences to be similar to the golden ratio or the pell numbers.
If you use the LambertW-function you will get around 1.02239 (I'm very exact since it concerns such a small number being impacted by 2 extremely different expressions).
How do you solve this please
The W-function is a type of logarithm that works for x×e^x. You know that ln(e^x)=x, the responding way of only getting x from x*e^x is writing W(x*e^x)=x. As long as both the factor and the exponent are identical you can use W successfully.
Even if you have (x/2)*e^(x/2) in an eqaution, you will get the factor in front of the exponential expression (being e^(x/2)). Thus, W((x/2)*e^(x/2))=x/2. With the fundementals understood, we can solve the equation in the video.
Basically, your goal in using the W-function is to get n*e^n=I, with n being any expression involving x and I being an integer.
2^x=x^32
(2^x)^1/32=(x^32)^1/32
2^(x/32)=x
1=x/(2^(x/32))
1=x*2^(-x/32)
1=x*(e^ln(2))^(-x/32)
1=x*e^(-(ln(2)*x/32))
-ln(2)/32=-ln(2)*x/32*e^(-ln(2)*x/32)
W(-ln(2)/32)=W(-ln(2)*x/32*e^(-ln(2)*x/32))
W(-ln(2)/32)=-ln(2)*x/32
W(-ln(2)/32)*32/(-ln(2))=x
Put this expression in Wolframalpha, instead of W, you write LambertW(numbers).
You're welcome, if you have difficulties understanding, then just ask.
put x=2^y, y=2^z, you get z=3, x = 2^(2^3) = 256. Done.
Giải hay lắm !❤❤❤
まさかのパワープレイw
Will somebody explain this to me how this works
애초에 지수식이랑 다항식인데 어떻게 연립을 하냐 ㅋㅋㅋ
Solve this problem or take a walk in the park ? Later y'all
I’m not bad at math I’m terrified of this song.
SUPERB! 😍😳🙃🌹🙏 MAI VREM! 🙏👏👍
5
vabb ma non è un metodo.. che ne sai quando trovi la combinazione giusta? potrebbe essere dopo 2 tentativi come dopo 1 miliardo. Si passa al log in base 2, poi applicando la formula del cambio di base si semplifica.
you can always brute force logs everywhere
Bueno estubo elegante el procediminto de base pero log sale en 2×3 rapido pero igislmente esta elegante
If you don't know how, just brute force it.
4 ane ki sambawna ha 😅