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.
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 ))
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.
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))
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
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.
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.
@@HandsomeHunk71 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)
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 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?
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.
@@jasonvuong2656its reasonable to assume that x is a power of 2. If it isn't we can try other stuff. Let x=2^y. 2^x=x^32 becomes 2^(2^y)=(2^y)^32. (2^y)^32 is the same as 2^(32y). Both sides have the same base, so the exponents must be equal. 2^y=32y. 32=2^5, so this can be rewritten as 2^(y-5)=y y must also be a power of 2, in order for y to be a factor of a power of 2. y=8 is the smallest power of 2 such that y-5 is a positive value, so it is an obvious first guess. y=8 works and that leads us to x=2^8=256.
Nice video to find the integer solution but if you graph those two function, the are more than one solution. there is at least one solution which is negative and real. But you have to use lambert W for that
It is rather "to guess", not "to solve"
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This why enterpreneurs are good on arithmetics but not good at algebra
czcams.com/video/ePVJLyHPnjQ/video.htmlsi=nIJMdBlYbU0npQMu
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...
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?
So what about x=1.022 and x=-0.979?
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.
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@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.
Pretty elegant solution! Thank you!
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Thanks for the interesting problem. I liked the way you present. Subscribing
Thank u dear sir❤️💖
I DONT UNSESTAND THE FINAL
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 ))
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
Where are the same bases ?
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.
A couple minutes of trial and error yielded x=1.0225 or thereabouts.
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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))
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i think you should show that the problem has only one solution to get all points, without that solution isn't full
А разве данное уравнение имеет не 3 решения? Автор указал всего 1, но их 3
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
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há mais de uma maneira de fazer, mas foi uma bela solução!!!
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It is like a endless math nightmare 😄 but thanks 👍
Very good thanks
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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 💕
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.
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.
Lovely
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2^256 = 1,1579*10^(77)
Икс ведь может быть ноль?
There's an identity in there somewhere, such as:
for a^x=x^b and a^n*b; x=b*n
Good
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
@@user-lx9hv9wl7x Yes same bro, but i can't find x= 256 in desmos plotting, why so? It shows only 2 intersections😢😢
@@HandsomeHunk71 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)
Nice Explaination Dear.❤
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@@K.Klogic my pleasure ☺️
Şöyle bi bakayım dedim denerken buldumn😅256 deneme yanılma cikiyor
Повезло, что показатель степени справа не 1024 😂
Why do you write like that?
En ese caso no era necesario decir? que X diferente de 0 y X > 0
using log is simplfy the equation
Pretty nice format, was easy to follow even without commentary
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You have lost two additional roots.
Música mto boa
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Full artificios 😮
there are 2 more solutions, x\approx -1.022 x\approx -0.979
why x 1/x
thx
Congratulations
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nice
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x = 256
I'm finally starting to solve problems like this 😊
Good dear sir 💖💕
Another answer is approximately x=1.023.
Good work
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.
-0.979 approximately next solution. There are three solutions
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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
whar is the name of the music used in video?
czcams.com/video/oeCvq6VbtmY/video.htmlsi=1wVFYPTyth1p9bxh
Это гениально
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Easy Question , you did it to difficulty.
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
WolframAlpha gave me 256 in less 1 second
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256 isn't the only solution.
Dear sir u can solve ❤️💖💕
Will the person even have the time left to finish the exam?? 😒😒
No dear sir ❤️💖
256,1.022,-0.979
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256
I solved this question using logarithms in 1 minute.
Good work dear❤️💖
5 al ojo
# 90 #
Ахуеть, и как я должен был это придумать?
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
I got X = 8 actually
I DONT UNDERSTAND THE FINAL
Watch again sir💕
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..
Где ещё один корень.? Должно быть два корня!
Same as first solution dear ❤️
Их три
x=256
naturelogrithium "ln"
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Bro this not a standard analytical way
Ok
2 1/32 =x 1/x
Значит x=x, значит 2=32 это абсурд)
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X=8
Noice
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Nunca entenderé las matemáticas 😔
Watch my videos one day u will understand
sireosly?😢
Yes
1
Thank you so much for ur comment
exist anoter solution
x is also approximately equal to 1.0223929402058
Why won't u explain and jus solving randomly
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Is n'nt anallytique nor mathamatical
# 45 #
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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.
x = x
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Бестолковые цифры. 99% никогда неиспользуют математику такую. Базовая нужна. А не вот такая порнушка
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Nu este o rezolvare coerenta.
Nu m- ai convins
Since it is obvious that x can only be a power of 2, simply by trying we see that 2^8 is a solution. That is actually faster way than yours ;).
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Can you show your work?
@@jasonvuong2656its reasonable to assume that x is a power of 2. If it isn't we can try other stuff. Let x=2^y.
2^x=x^32 becomes 2^(2^y)=(2^y)^32.
(2^y)^32 is the same as 2^(32y).
Both sides have the same base, so the exponents must be equal.
2^y=32y.
32=2^5, so this can be rewritten as 2^(y-5)=y
y must also be a power of 2, in order for y to be a factor of a power of 2.
y=8 is the smallest power of 2 such that y-5 is a positive value, so it is an obvious first guess.
y=8 works and that leads us to x=2^8=256.
@@jakefromstatefarm6969 my friend, your method is way more complicated than the guy doing the clip.
@@jasonvuong2656 part of that is due to text based limitations. It's quicker, but I can see how it is more complicated.
nice problem
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Nice video to find the integer solution
but if you graph those two function, the are more than one solution. there is at least one solution which is negative and real. But you have to use lambert W for that
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$100 million question, where is it in nature? 😂
Where are you in nature ?
You were supposed to apply a logarithm not that nonsense
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VERY VERY WRONG
Ok
very childish writing, lot’s of unnecessary talks
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Perfect by-pass 👌
Thank u❤️
this problem have 3 answer