This method isn’t analytical. For instance, it doesn’t answer if this equation has only one solution. For the similar equations like that but with different coefficients it will not work.
Take a guess and assume that like 32, the answer is a power of two. Use logarithms in base 2 because log(2, 2^x) = x. and log(2, x^32) = 32 log(2, x). Or, x = 32 log(2, x). So x / log(2,x) = 32. This can be solved by inspection, again beginning with powers of 2: x is divided by x as some power of 2, Trying a few values of x we see that log(2,256) = 8 and 256/8 = 32.
It may be worth noting that 2 divides 2^x if x is natural number, so 2 would have to divide the right hand side as well. I think it's Euclid's Lemma that would then tell us x has to be at least a multiple of 2, if not a power of 2.
Take log two of both sides then you get X equals log base two of x multiplied by 32 other words what number when you take it’s logarithm it does the same thing as dividing it by 32. You can make educated guesses until you stubble upon 256.
Это не совсем подбор. Методология и тактика тут есть. Подбор был сделан под конец, что подпортило эффект. Подбора можно было и не делать. Но понадобилась бы вычислит машина. И снова была б критика ))
Если условие найти натуральный корень, то решение класс, если количество корней, то по графику видно два и третий, который 256, его надо догадаться, за интересное уравнение спасибо!
x^(1/x) is a monotone decreasing function if x>e, then the equation x^(1/x)=2^(1/32) has at most one real solution. x=256 works, so no other real soultion.
I use the Lambert W function to resolve x^32=2^x. «ln» both sides : 32 lnx= x ln2. So lnx/x = ln2/32 the same as (x^-1) lnx = ln2/32. x^-1 = e(ln(x^-1)) and lnx = -ln(x^-1). So we have : e(lnx^-1) x ln(x^-1)= -ln2/32. So lnx^-1 = W(-ln2/32). «e» both sides : x = 1/(e(W(-ln2/32))
Если расписать левую и правую часть как функции, то за счёт чётности, функция f(x)=X^32 будет пересекать функцию g(x)=2^x в двух точках: одно решение потеряно. P. S. ради интереса забил в Desmosе эти 2 функции и абсциссы их пересечений равны 1.022 и -0.979. Вот так вот интересно.
@@Swamisonic18 Most likely, this is due to the fact that the polynomial of the nth degree has at most n roots. In our case, this is a polynomial of the 32nd degree on the right side, which essentially limits the number of roots of this equation. That is, in this equation it has already been proved that there are at least 3 real roots, and all the others are complex. (Unless we find other real roots)
If you need to resort to manipulating in the way that you did, I think switching to logarithms is far easier. You simply have to find a multiple of 32 that includes as a factor a log to the base of 2 that generates itself....namely 256, which is 32 times log to the base 2 of, you guessed it, 256. If the kids are in the Olympiad selection process, that kind of arithmetic via logs is child's play to them.
The difference between \(2^{1.0224}\) and \((1.0224)^{32}\) is approximately \(-0.00044\). This small discrepancy indicates that \(1.0224\) is a very close approximation to the solution of the equation \(2^x = x^{32}\). ChatGPT
Using logarithmic fuction is the true solution for this particular sum...because it is not a solution unless you are able to solve all the similar types of sums using that method, irrespective of the base and the exponent...for example if it was 31 instead of 32...using this method would have fetched us nothing at all.
TZZ {4,2,1} Tzz is the Taha + Collatz brand. Factories of T-shirt, sports shoes, engineering machinery and tools, airplanes, cars, trains and ships must use this mighty sign that signifies victory and the solution of Collatz. So factories must agree with me to use this mark. Author: Taha M. Muhammad/ USA Kurd Iraq Bakuage Thank you prepare your prize
But Why we don’t do any not equivalent transformation? Or do? To loose another solution in real numbers? Or there is negative solutions have been lost?
Es una solución para estos exponentes con otros exponentes no funciona, no es un método solo es una solución, con logaritmos es más fácil resolver este tipo de ecuaciones.
32-довольно большая степень, чтобы на графике рассмотреть точку пересечения при x=256, поэтому для начала посмотрите на пересечения графиков y=2^x и y=x^2
maybe you can say x=2^k (where x>0) and get 2^k=32.k or f(k)=2^k-32k anf f'=lnk.2^k-32 than pick a root, lets say k=1 and apply newton-raphson algorithm, probably in a few steps you will get a very close numerical solution. for negative root you can say x=-2^k (x
@@jeanpierre7971 yes, there's no guarantee for a unique solution If it's bijective, it has a unique sol What I meant is that x -> x^(1/x) does not need to be bijective to make one solution works as long as there is a verification at the end
Given that this is a competition question posed to us, there is no need to prove anything. All that is needed is determining a solution, not proving that it is a unique solution.
It is rather "to guess", not "to solve"
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This why enterpreneurs are good on arithmetics but not good at algebra
ua-cam.com/video/ePVJLyHPnjQ/v-deo.htmlsi=nIJMdBlYbU0npQMu
there is something called “mathematical induction”
it is legit
這確實不是好的解答
This method isn’t analytical. For instance, it doesn’t answer if this equation has only one solution. For the similar equations like that but with different coefficients it will not work.
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Indeed, there are 2 other real solutions, plus multiple families of complex solutions.
@@jursamaj1 more real...
Есть и проще решение чем то что в видео. Найти решение на пересечении графиков функций. Вот и все.
Thanks for the interesting problem. I liked the way you present. Subscribing
Thank u dear sir❤️💖
Just use logarithm
Geterosexual logarithm*
You would receive something with the Lambert W function (irrational in general)
mу соmmеnt wаs dеlеtеd....
It sаid:
hеtеrоsехuаl lоgаrithm
@@progr6171спалился
@@progr6171
And is there the homosexual one?
Take a guess and assume that like 32, the answer is a power of two. Use logarithms in base 2 because log(2, 2^x) = x. and log(2, x^32) = 32 log(2, x). Or, x = 32 log(2, x). So x / log(2,x) = 32. This can be solved by inspection, again beginning with powers of 2: x is divided by x as some power of 2, Trying a few values of x we see that log(2,256) = 8 and 256/8 = 32.
Thank you for your comment ❤️💖
@markharder3676 Thank you.
This method for people who don’t know logarithm
Very convoluted
@@diwakaranvalangaimanmani3777if you know exponents you have to know logaritms😂
It may be worth noting that 2 divides 2^x if x is natural number, so 2 would have to divide the right hand side as well. I think it's Euclid's Lemma that would then tell us x has to be at least a multiple of 2, if not a power of 2.
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Doesn't look like anyone is willing to solve the problem, just guess at a solution! What if the problem was 2^x = x^31?
Then X would equal 1.0231407...
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@@dhy5342 or about 246.266
@@dhy5342 Also 246.266…, plus 5 families of complex answers.
Take log two of both sides then you get X equals log base two of x multiplied by 32 other words what number when you take it’s logarithm it does the same thing as dividing it by 32. You can make educated guesses until you stubble upon 256.
Сложно назвать это математическим методом решения, скорее простой подбор.
Это не совсем подбор.
Методология и тактика тут есть.
Подбор был сделан под конец, что подпортило эффект. Подбора можно было и не делать. Но понадобилась бы вычислит машина. И снова была б критика ))
@@elmanmusaiev7003 зачем машина, выше правильно про логарифмы написали. А показаний способ это все же подбор.
A couple minutes of trial and error yielded x=1.0225 or thereabouts.
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A matemática é um jogo fascinante!!! Gostei.
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Если условие найти натуральный корень, то решение класс, если количество корней, то по графику видно два и третий, который 256, его надо догадаться, за интересное уравнение спасибо!
Thanks 💕
i think you should show that the problem has only one solution to get all points, without that solution isn't full
x^(1/x) is a monotone decreasing function if x>e, then the equation
x^(1/x)=2^(1/32) has at most one real solution. x=256 works, so no other real soultion.
I DONT UNSESTAND THE FINAL
I use the Lambert W function to resolve x^32=2^x.
«ln» both sides : 32 lnx= x ln2.
So lnx/x = ln2/32 the same as (x^-1) lnx = ln2/32.
x^-1 = e(ln(x^-1)) and lnx = -ln(x^-1).
So we have : e(lnx^-1) x ln(x^-1)= -ln2/32. So lnx^-1 = W(-ln2/32).
«e» both sides : x = 1/(e(W(-ln2/32))
Thank u for your solution ❤️💕💖
So what about x=1.022 and x=-0.979?
Error
It is like a endless math nightmare 😄 but thanks 👍
Pretty elegant solution! Thank you!
Thanks for your helpful comment ❤️
It looks to me as if there must be a solution between x=1 and x=2
Ok
@@K.Klogicand a negative solution. There are 3 real solutions.
How to find all solutions? Tell the method please🙏@@rubixpuzzlechamp
Curves 2^x and x^32 intersect twice: x at one intersection is positive, and at the second intersection is negative. So there should be two solutions.
Good dear ❤️
Nice Explaination Dear.❤
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@@K.Klogic my pleasure ☺️
Если расписать левую и правую часть как функции, то за счёт чётности, функция f(x)=X^32 будет пересекать функцию g(x)=2^x в двух точках: одно решение потеряно.
P. S. ради интереса забил в Desmosе эти 2 функции и абсциссы их пересечений равны 1.022 и -0.979. Вот так вот интересно.
Good work
Думаю, что не ординаты, а абсциссы, но их три, еще 256
@@user-vs7it4bc6k Спасибо, исправил
@@Alex_the_engineer737 Yes same bro, but i can't find x= 256 in desmos plotting, why so? It shows only 2 intersections😢😢
@@Swamisonic18 Most likely, this is due to the fact that the polynomial of the nth degree has at most n roots. In our case, this is a polynomial of the 32nd degree on the right side, which essentially limits the number of roots of this equation. That is, in this equation it has already been proved that there are at least 3 real roots, and all the others are complex. (Unless we find other real roots)
If you need to resort to manipulating in the way that you did, I think switching to logarithms is far easier. You simply have to find a multiple of 32 that includes as a factor a log to the base of 2 that generates itself....namely 256, which is 32 times log to the base 2 of, you guessed it, 256. If the kids are in the Olympiad selection process, that kind of arithmetic via logs is child's play to them.
The difference between \(2^{1.0224}\) and \((1.0224)^{32}\) is approximately \(-0.00044\). This small discrepancy indicates that \(1.0224\) is a very close approximation to the solution of the equation \(2^x = x^{32}\).
ChatGPT
Thank you for your comment ❤️
This type of problem is still big problem fot gpt but it can solve it in the end, wiil wait fifth gen.
А разве данное уравнение имеет не 3 решения? Автор указал всего 1, но их 3
há mais de uma maneira de fazer, mas foi uma bela solução!!!
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2^256 = 1,1579*10^(77)
Cevap 1 oluyor 2 üzeri 5 x in üzeri 32 den x le 2uzeri 5 yer değiştirme yaparak üste 5 üzeri x kalır oda x sıfır dan 2 ye eşitlenir.
Where are the same bases ?
Using logarithmic fuction is the true solution for this particular sum...because it is not a solution unless you are able to solve all the similar types of sums using that method, irrespective of the base and the exponent...for example if it was 31 instead of 32...using this method would have fetched us nothing at all.
By taking log this will provide trancedental equations to be solved graphically
Good
What about the other two solutions
This is guess and check method, so you must prove there are no other solutions.
Exceptional
Very good thanks
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Решается переобозначением x = 2^y, а затем простым подбором.
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I must be super drunk off my ass to have watched the whole thing
Pretty nice format, was easy to follow even without commentary
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Great video,
Sir how can we communicate with you regarding to any misunderstanding sie
How can I help you?
You have lost two additional roots.
Şöyle bi bakayım dedim denerken buldumn😅256 deneme yanılma cikiyor
x = 256
I'm finally starting to solve problems like this 😊
Good dear sir 💖💕
Sir if there is any G chat available? Through gmail so can discuss further regarding to topic
Thanks for the video. Maybe someone knows how to find these two more solutions for this problem? (~ -0.979 and 1.0225)
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Yes, I published actual solution in comments
using log is simplfy the equation
I don't mind the method, but do mind the time it took
Lovely
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Повезло, что показатель степени справа не 1024 😂
Log kis din Kam ayega Bhai
Another answer is approximately x=1.023.
Good work
En ese caso no era necesario decir? que X diferente de 0 y X > 0
There's an identity in there somewhere, such as:
for a^x=x^b and a^n*b; x=b*n
Good
there are 2 more solutions, x\approx -1.022 x\approx -0.979
x=2^n=32n
n=2^(n-5)=8
x=2^8=256
256,1.022,-0.979
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Икс ведь может быть ноль?
-0.979 approximately next solution. There are three solutions
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TZZ
{4,2,1}
Tzz is the Taha + Collatz brand. Factories of T-shirt, sports shoes, engineering machinery and tools, airplanes, cars, trains and ships must use this mighty sign that signifies victory and the solution of Collatz. So factories must agree with me to use this mark.
Author: Taha M. Muhammad/ USA Kurd Iraq
Bakuage Thank you prepare your prize
But you don't proof that is only one solution. In naturals, yes, one 256, but in reals is one more root at interval [1,2]
But Why we don’t do any not equivalent transformation? Or do? To loose another solution in real numbers?
Or there is negative solutions have been lost?
@@anton-ke4qzThis solution is not correct
This equation has infinite answers. That's only one of the answers or roots
The answer is not complete. I guess there is another solution in (-1,0) interval. And another one in (1, 2) interval.
Yes dear i mention already
Imma try it.
whar is the name of the music used in video?
ua-cam.com/video/oeCvq6VbtmY/v-deo.htmlsi=1wVFYPTyth1p9bxh
Easy Question , you did it to difficulty.
Congratulations
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Too many steps. Waste of time.
Es una solución para estos exponentes con otros exponentes no funciona, no es un método solo es una solución, con logaritmos es más fácil resolver este tipo de ecuaciones.
But there is a second solution: X = 1,0223929... How can I find this second solution?
Это неверное решение
@@-Skynet- Постройте графики функций y=2^x и y=x^32 и посмотрите где они пересекаются.
x≈−0.97901693,1.02239294, 256
32-довольно большая степень, чтобы на графике рассмотреть точку пересечения при x=256, поэтому для начала посмотрите на пересечения графиков y=2^x и y=x^2
@@-Skynet- Проверку проходит?
🙂
maybe you can say x=2^k (where x>0) and get 2^k=32.k or f(k)=2^k-32k anf f'=lnk.2^k-32
than pick a root, lets say k=1 and apply newton-raphson algorithm, probably in a few steps you will get a very close numerical solution.
for negative root you can say x=-2^k (x
Вот мне в жизни нигде это не пригодилось(((
Música mto boa
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Это гениально
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Cant we use log here?
Try first
Will the person even have the time left to finish the exam?? 😒😒
No dear sir ❤️💖
Why do you write like that?
I solved this question using logarithms in 1 minute.
Good work dear❤️💖
Full artificios 😮
256 isn't the only solution.
Dear sir u can solve ❤️💖💕
Где ещё один корень.? Должно быть два корня!
Same as first solution dear ❤️
Их три
WolframAlpha gave me 256 in less 1 second
Great ❤️💖💕
nice
Thanks for your helpful comment ❤️
X=1. Prove me wrong
You are wrong
2^1=2
1^32=1
2 1/32 =x 1/x
Значит x=x, значит 2=32 это абсурд)
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I love the way you communicate without words.
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...and with decent music, unlike so many other channels. I have just liked and subscribed.
@@davidbrown8763 thank you dear sir.supor my channel thank u💖❤️🎉
@@K.Klogic I am a supporter.
@@davidbrown8763 thank you dear..
thx
why x 1/x
I DONT UNDERSTAND THE FINAL
Watch again sir💕
Don't you need to prove that x -> x^(1/x) is bijective?
not really as long as you do the verification at the end
@@Drestanto Then what guarantees that there is no other solution?
@@jeanpierre7971 yes, there's no guarantee for a unique solution
If it's bijective, it has a unique sol
What I meant is that x -> x^(1/x) does not need to be bijective to make one solution works as long as there is a verification at the end
Given that this is a competition question posed to us, there is no need to prove anything.
All that is needed is determining a solution, not proving that it is a unique solution.
@@jakefromstatefarm6969 You are very wrong. First of all, there are two solutions though only one is an integer.
Bro this not a standard analytical way
Ok
❤
naturelogrithium "ln"
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sireosly?😢
Yes
x=256
Nunca entenderé las matemáticas 😔
Watch my videos one day u will understand
Ахуеть, и как я должен был это придумать?
Mashaallah,
밑2인 로그를 써라
256
Noice
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No sé vio como resolvió al final
Watch again