Germany - Math Olympiad Exponential Problem.

Germany - Math Olympiad Exponential Problem.

[@maciejterakowski9062]

It is rather "to guess", not "to solve"

[@K.Klogic]

Thank you for your comment ❤️💖

[@jb.drills]

This why enterpreneurs are good on arithmetics but not good at algebra

[@mathophilechirag]

kzbin.info/www/bejne/m4G5e3-vfbWhoLMsi=nIJMdBlYbU0npQMu

[@vvv1041]

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.

[@K.Klogic]

Thank you for your comment ❤️💖

[@jursamaj]

Indeed, there are 2 other real solutions, plus multiple families of complex solutions.

[@sergeykardash4482]

​​@@jursamaj1 more real...

[@howardhavlicsek5031]

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?

[@dhy5342]

Then X would equal 1.0231407...

[@K.Klogic]

Thank you for your comment ❤️

[@K.Klogic]

Thank you for your comment ❤️💖

[@aR3DFgH8kLj2pOqIuVx5]

@@dhy5342 or about 246.266

[@jursamaj]

@@dhy5342 Also 246.266…, plus 5 families of complex answers.

[@timmy18135]

Just use logarithm

[@progr6171]

Geterosexual logarithm*

[@saturninufolud6767]

You would receive something with the Lambert W function (irrational in general)

[@progr6171]

mу соmmеnt wаs dеlеtеd.... It sаid: hеtеrоsехuаl lоgаrithm

[@gmddlex5873]

​@@progr6171спалился

[@AW-dt8ct]

​​@@progr6171 And is there the homosexual one?

[@markharder3676]

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.

[@K.Klogic]

Thank you for your comment ❤️💖

[@rickkwitkoski1976]

@markharder3676 Thank you.

[@diwakaranvalangaimanmani3777]

This method for people who don’t know logarithm

[@b213videoz]

Very convoluted

[@antoniopetrolagarrigos9771]

​@@diwakaranvalangaimanmani3777if you know exponents you have to know logaritms😂

[@aleksandrajones4101]

So what about x=1.022 and x=-0.979?

[@Limited_Light]

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.

[@K.Klogic]

Thank you for your comment ❤️

[@JSSTyger]

It looks to me as if there must be a solution between x=1 and x=2

[@K.Klogic]

Ok

[@rubixpuzzlechamp]

@@K.Klogicand a negative solution. There are 3 real solutions.

[@HandsomeHunk71]

How to find all solutions? Tell the method please🙏​@@rubixpuzzlechamp

[@pub7496]

Thanks for the interesting problem. I liked the way you present. Subscribing

[@K.Klogic]

Thank u dear sir❤️💖

[@jamesmarshall7756]

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))

[@K.Klogic]

Thank u for your solution ❤️💕💖

[@EasyMathSteps-SolveItEasy]

Pretty elegant solution! Thank you!

[@K.Klogic]

Thanks for your helpful comment ❤️

[@chrisj5443]

A couple minutes of trial and error yielded x=1.0225 or thereabouts.

[@K.Klogic]

Thank you for your comment ❤️

[@DumAndSmart]

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.

[@EliabBezerrademello]

I DONT UNSESTAND THE FINAL

[@hkleiser5848]

i think you should show that the problem has only one solution to get all points, without that solution isn't full

[@jamesmarshall7756]

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

[@K.Klogic]

Thank you for your comment ❤️

[@77das]

Сложно назвать это математическим методом решения, скорее простой подбор.

[@elmanmusaiev7003]

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 ))

[@ArinaAnn]

​@@elmanmusaiev7003 зачем машина, выше правильно про логарифмы написали. А показаний способ это все же подбор.

[@user-lx9hv9wl7x]

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.

[@K.Klogic]

Good work

[@user-vs7it4bc6k]

Думаю, что не ординаты, а абсциссы, но их три, еще 256

[@user-lx9hv9wl7x]

@@user-vs7it4bc6k Спасибо, исправил

[@HandsomeHunk71]

​@@user-lx9hv9wl7x Yes same bro, but i can't find x= 256 in desmos plotting, why so? It shows only 2 intersections😢😢

[@user-lx9hv9wl7x]

@@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)

[@nenemtiaof5016]

Where are the same bases ?

[@richardrakamazi5754]

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.

[@recepgurkanhosman155]

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.

[@Observer1973]

You have lost two additional roots.

[@alexeyevgenevich9821]

2^256 = 1,1579*10^(77)

[@user-vs7it4bc6k]

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!

[@K.Klogic]

Thanks 💕

[@Mb-logic]

Nice Explaination Dear.❤

[@K.Klogic]

Thank you for your comment ❤️💖

[@Mb-logic]

@@K.Klogic my pleasure ☺️

[@user-hy4em8tf7e]

А разве данное уравнение имеет не 3 решения? Автор указал всего 1, но их 3

[@louf7178]

There's an identity in there somewhere, such as: for a^x=x^b and a^n*b; x=b*n

[@K.Klogic]

Good

[@AzmiTabish]

Lovely

[@K.Klogic]

Thank u dear for your comment ❤️💖

[@etc82]

It is like a endless math nightmare 😄 but thanks 👍

[@user-yc4wh8lh8b]

Very good thanks

[@K.Klogic]

Thanks for your comment ❤️💖

[@Songfugel]

Pretty nice format, was easy to follow even without commentary

[@K.Klogic]

Thank you for your comment ❤️

[@user-bp7zr1nu3y]

Easy Question , you did it to difficulty.

[@user-iw8yh1ib4n]

Thanks for the video. Maybe someone knows how to find these two more solutions for this problem? (~ -0.979 and 1.0225)

[@K.Klogic]

Thank you for your comment ❤️

[@screemer949]

Yes, I published actual solution in comments

[@MrTaiado]

há mais de uma maneira de fazer, mas foi uma bela solução!!!

[@K.Klogic]

Thank you for your comment ❤️

[@hectoraular6180]

En ese caso no era necesario decir? que X diferente de 0 y X > 0

[@radutrimbitas4930]

there are 2 more solutions, x\approx -1.022 x\approx -0.979

[@vasiliynikiforov1976]

Икс ведь может быть ноль?

[@YasinBEY-eo1ng]

Şöyle bi bakayım dedim denerken buldumn😅256 deneme yanılma cikiyor

[@laith.alzubaidi]

using log is simplfy the equation

[@user-oc6yd4vy6z]

Another answer is approximately x=1.023.

[@K.Klogic]

Good work

[@DuyTran-uj1cv]

why x 1/x

[@user-xe1cr1ni5x]

-0.979 approximately next solution. There are three solutions

[@K.Klogic]

Thank you for your comment ❤️💖

[@glitch9643]

Cant we use log here?

[@K.Klogic]

Try first

[@user-du5er5je7i]

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]

[@anton-ke4qz]

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?

[@user-du5er5je7i]

​@@anton-ke4qzThis solution is not correct

[@gaznelimahmut1769]

thx

[@AlfredoMartinez-hm2tb]

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.

[@sidneiroberto3894]

Música mto boa

[@K.Klogic]

Thank you for your comment ❤️💖

[@DumAndSmart]

256,1.022,-0.979

[@K.Klogic]

Thank you for your comment ❤️💖

[@aizaypubg5956]

nice

[@K.Klogic]

Thanks for your helpful comment ❤️

[@jhoanuceda8145]

Full artificios 😮

[@user-ob4dc4co5l]

Повезло, что показатель степени справа не 1024 😂

[@mysoultrade]

Это гениально

[@K.Klogic]

Thank u dear for your comment ❤️

[@agniksen8507]

Will the person even have the time left to finish the exam?? 😒😒

[@K.Klogic]

No dear sir ❤️💖

[@cosmosapien597]

Why do you write like that?

[@pranshuprajnabehera139]

whar is the name of the music used in video?

[@eliascuba420]

kzbin.info/www/bejne/pZamp6Rsi8eno7ssi=1wVFYPTyth1p9bxh

[@b213videoz]

x = 256 I'm finally starting to solve problems like this 😊

[@K.Klogic]

Good dear sir 💖💕

[@ziyodabekmirzayeva9240]

Mashaallah,

[@EliabBezerrademello]

I DONT UNDERSTAND THE FINAL

[@K.Klogic]

Watch again sir💕

[@joseagra7725]

Congratulations

[@K.Klogic]

Thank you for your comment ❤️💖

[@stevenschmidt]

256 isn't the only solution.

[@K.Klogic]

Dear sir u can solve ❤️💖💕

[@wwxyz7570]

WolframAlpha gave me 256 in less 1 second

[@K.Klogic]

Great ❤️💖💕

[@lool9935]

Bro this not a standard analytical way

[@K.Klogic]

Ok

[@mcrow312166]

I love the way you communicate without words.

[@K.Klogic]

Thank you for your comment ❤️

[@davidbrown8763]

...and with decent music, unlike so many other channels. I have just liked and subscribed.

[@K.Klogic]

@@davidbrown8763 thank you dear sir.supor my channel thank u💖❤️🎉

[@davidbrown8763]

@@K.Klogic I am a supporter.

[@K.Klogic]

@@davidbrown8763 thank you dear..

[@jewi71]

But there is a second solution: X = 1,0223929... How can I find this second solution?

[@-Skynet-]

Это неверное решение

[@alexanderfishfinder]

@@-Skynet- Постройте графики функций y=2^x и y=x^32 и посмотрите где они пересекаются. x≈−0.97901693,1.02239294, 256

[@alexanderfishfinder]

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

[@user-mn5wu7xx4d]

@@-Skynet- Проверку проходит? 🙂

[@Sabri1729]

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

[@user-kd8gn3zu6f]

naturelogrithium "ln"

[@K.Klogic]

Thank you for your comment ❤️💖

[@lamervrte1482]

I solved this question using logarithms in 1 minute.

[@K.Klogic]

Good work dear❤️💖

[@katabeth8272]

X=8

[@user-qy8re3yx3d]

x=256

[@DimaCharapko]

sireosly?😢

[@K.Klogic]

Yes

[@user-iw2zn2mj3c]

Где ещё один корень.? Должно быть два корня!

[@K.Klogic]

Same as first solution dear ❤️

[@screemer949]

Их три

[@samiqala6543]

256

[@ouldmedouldmedmr9489]

Is n'nt anallytique nor mathamatical

[@imranfahami5694]

I got X = 8 actually

[@tamarshahverdyan2723]

# 90 #

[@Jeff_D]

5 al ojo

[@jeanpierre7971]

Don't you need to prove that x -> x^(1/x) is bijective?

[@Drestanto]

not really as long as you do the verification at the end

[@jeanpierre7971]

@@Drestanto Then what guarantees that there is no other solution?

[@Drestanto]

@@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

[@jakefromstatefarm6969]

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.

[@pedjolinko]

@@jakefromstatefarm6969 You are very wrong. First of all, there are two solutions though only one is an integer.

[@yatingkang3257]

x is also approximately equal to 1.0223929402058

[@user-sg1dk3mf7d]

2 1/32 =x 1/x Значит x=x, значит 2=32 это абсурд)

[@K.Klogic]

Thank u for your comment ❤️💖

[@tiporocstar1106]

Nunca entenderé las matemáticas 😔

[@K.Klogic]

Watch my videos one day u will understand

[@lucianopires1184]

exist anoter solution

[@theophilusprat2202]

Why won't u explain and jus solving randomly

[@K.Klogic]

❤️

[@nikp941]

Noice

[@K.Klogic]

Thanks for your comment ❤️💖

[@Novomokoroni]

Ахуеть, и как я должен был это придумать?

[@tamarshahverdyan2723]

# 45 #

[@K.Klogic]

Thank you for your comment ❤️💖

[@K.Klogic]

Thank you for your comment ❤️💖

[@user-bn2gl3cu4v]

Nu este o rezolvare coerenta. Nu m- ai convins

[@perakojot6524]

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 ;).

[@K.Klogic]

Thank u for your comment ❤️❤️❤️

[@jasonvuong2656]

Can you show your work?

[@jakefromstatefarm6969]

​@@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.

[@jasonvuong2656]

@@jakefromstatefarm6969 my friend, your method is way more complicated than the guy doing the clip.

[@jakefromstatefarm6969]

@@jasonvuong2656 part of that is due to text based limitations. It's quicker, but I can see how it is more complicated.

[@user-kg5rr9rv2g]

1

[@K.Klogic]

Thank you so much for ur comment

[@yvindmathiassen7617]

nice problem

[@K.Klogic]

Thank you for your comment ❤️

[@stighenningjohansen]

x = x

[@K.Klogic]

Thank you for your comment ❤️💖

[@Tatarin374]

Бестолковые цифры. 99% никогда неиспользуют математику такую. Базовая нужна. А не вот такая порнушка

[@K.Klogic]

Respect your comment ❤️

[@Drestanto]

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

[@K.Klogic]

Thank you for your comment ❤️

[@Lincoln-ht2dv]

Too many steps. Waste of time.

[@Arrotte.Weiler]

$100 million question, where is it in nature? 😂

[@trinanjanchatterjee2092]

Where are you in nature ?

[@user-ot7xu8cc3l]

You were supposed to apply a logarithm not that nonsense

[@K.Klogic]

I respect your comment ❤️

[@itutixnoti8623]

VERY VERY WRONG

[@K.Klogic]

Ok

[@yanfisher2639]

very childish writing, lot’s of unnecessary talks

[@K.Klogic]

I respect your comment ❤️

[@elmanmusaiev7003]

Perfect by-pass 👌

[@K.Klogic]

Thank u❤️

[@hungzz69]

this problem have 3 answer