Germany - Math Olympiad Exponential Problem

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[@dmitrymelnik8296]

The easier solution is to do substitution x --> 2^m. Then, we get 2^m = 32m, or m = 2^(m-5), hence m is also a power of two (if we are concerned only with the integer solutions). The smallest value m=8 satisfies the equation. hence m=8, x=2^8 = 256.

[@jkr9224]

Those interested in maths will take interest in learning this 🎉🎉🎉🎉🎉🎉 Superb 🎉

[@sudhangshubhattacharya4991]

Very nicely proceeding towards the target

[@yapadek3098]

Hi, and thank you for your videos. How can you be sure this solution, obtainend by identification (is that the word ?) is unique ?

[@VoiceTuned]

Wow very nice

[@Eleuthero5]

Wow that was a very entertaining proof!!

[@learncommunolizer]

Thank you very much!

[@Ibrahimcubesolevtaknik]

Very good solution

[@AKRiitMedPhysicsGuru]

Nice bro🎉

[@hoahocphothongmoi2212]

very good. Thanks you!!!!!!!!

[@alejomdp]

Your answer is right but when you analize the graphic of 2^x and x^32 the graphic of both functions intersects at three points: x ≈ -0.979, x ≈ 1.022 AND x = 256. For those first two solutions I couldn't find an analytical way to calculate their exact value. I know they are irrational but I couldn't find a way to express them with roots or even something with the number e.

[@PIANOJOE7]

can´t be solved analytically

[@albedogray]

@@PIANOJOE7why?

[@albedogray]

Also realized that the equation has 3 roots, can't find an analytical solution (

[@rslua01]

Ty

[@namantiwari4529]

Finnally bro did it

[@nefando7777]

Me encanta cómo dice "two".

[@WenjieGu]

Très intéressant

[@learncommunolizer]

Thank you very much !

[@richardleveson6467]

Very nice! I wondered where you were going but suddenly you arrived.

[@muhammadarshadh633]

That's what she said

[@user-ki7im6ym6n]

just use the logarithm equation

[@gregsti]

Très intéressant !

[@learncommunolizer]

Thank you very much!

[@claudebalzano7031]

Cette méthode pose le problème de la bijectivité de la fonction x^(1/x), non ?

[@user-pi2my7vk5j]

Интересное задание и его решение!

[@PacoMoyajaquemate]

El problema tiene 3 soluciones: x= 1.02239, x= -0.97902

[@SVC_Editz]

Also 1.0223

[@oliverdauphin236]

Méthode intuitive. On remarque que 2^8=256 et que 8*32=256 Du coup on suppose x=2^8 puis on vérifie : 2^256=(2^8)*32=2^(8*32)=2^256

[@user-sy4tf6rv8l]

Thanks

[@julianbruns7459]

Unless x was restricted to integers, i would rate this answer as incomplete. Although there is only 1 integer solution, there are 3 real solutions (and an infinite amount of complex solutions). Did i miss the part where it was specified what set x belongs to?

[@user-em7rz8kk4r]

xは多分2^tと書ける、指数部分比較して2^t=32tを解く→t=8からx=256

[@ayugaming3047]

Could you please clarify my brother? The steps ...

[@user-em7rz8kk4r]

@@ayugaming3047 First of all I thought the answer to this question could be expressed as 2^t. Then I substituted and compared the exponential part: 2^t=32t and 2^t is monotonically increasing, so there is one solution. So t=8. Sorry for my poor English

[@ayugaming3047]

@@user-em7rz8kk4r your English is fine. So, what calculations did you use after the step 2^x = 32t? Did you just guess?

[@user-em7rz8kk4r]

@@ayugaming3047 Mental arithmetic

[@ayugaming3047]

@@user-em7rz8kk4r can you introduce that to me?

[@pianoplayer123able]

The other solution is: 1/e^lambertw(-(ln(2)/32))≈1.0223

[@allozovsky]

But there should also be a negative real solution, since (−x)³² = x³² and 2ˣ = (1/2)⁻ˣ, so we get an equation (1/2)⁻ˣ = (−x)³² for −x ∈ ℝ.

[@dmitrymelnik8296]

@@allozovsky Functions 2^x and x^32 intersect only in two points.

[@allozovsky]

You may plot them on the interval from −1.5 to 1.5 and find out that they indeed intersect at two non-integer points (and a third one at x = 256).

[@dmitrymelnik8296]

@@allozovsky yeah, my bad.

[@allozovsky]

It's alright. Solving for irrational roots requires application of the Lambert W function or using numerical methods, so they can't be found by elementary algebraic transformations anyway.

[@huseyinaydogan6928]

this answer is imcomplete. there are infinite amount of solutions.(x=256 is not the only one)

[@albedogray]

why infinite?

[@oliverdauphin236]

2^x= x^32 Posons x=2^n L èquatiion devient 2^(2^n)=(2^n)^32 Donc 2^n=32n en identifiant les exposants Posons n=2^a L’équation devient 2^(2^a}=(2^5)*(2^a) Donc 2^(2^a}=2^(5+a) Donc 2^a=5+a en identifiant les exposants Or 2^3=8=5+a Donc a=3 Donc n=2^3=8 Donc x=2^8=256 Donc 2^256=256^32 On vérifie : 2^256=(2^8)*32=2^(8*32)=2^256

[@Ha_Gia_2022]

Smart solving ! Thanks. Just last eq. 2^a = (a+5) how to obtain a=3 ?

[@user-qq9sq7pk5k]

X. 2^x. X^32 0. 1. > 0 1. 2. > 1 2. 4.

[@dinlendiricidrtv]

Because 256

[@salehabasov7074]

Azərbaycancan salamlar niqırr(jholooom) Alqasımlıdan salamlar helom məllim😂😂😂😂

[@vishnujatav6329]

Tricky

[@xgx899]

It is bad enough that only one solution is found. What is worse, is that uniqueness is not even discussed. Why so many people like to lecture on mathematics without first learning the basics of ``mathematical culture".

[@Mralipol]

Bir ile iki arasında bir x daha olsa gerek onunda bulabilir misiniz?

[@Rao_adib]

It was a lengthy solution. Rather it can be solved by taking natural log on both sides

[@ivangorin1254]

В этом уравнении имеется 3 корня -0,97902 1,022393 256

[@Austin101123]

This needs lambert W function to find 3 real solutions (infinite complex ones).

[@SouvereineMaman]

Not Bad like methode

[@learncommunolizer]

Thank you very much!!!

[@himadrikhanra7463]

(1)^x/32=2^32 ×x...so x= 1/2^32....

[@ConvivialWorldTravel]

5^x = x^15, Can it solve same way? pls help. Thanks

[@miracbarsapaydn650]

X = 256

[@kennycovenas3598]

Muy complejo el procedimiento

[@user-sv2hi6ku8c]

Hola cómo estás todo ustedes bueno día desde san Felipe de Puerto plata primera espalda de la restauración general Gregorio luperon machete carajo 🇩🇴⚔️☕

[@alfredteichmann135]

Take two sheets of paper!

[@pujarian]

Just take log both sides

[@nahideipekcioglu7552]

Çözüm cok uzun malesef.🤷‍♀️ 2 nin kuvvetlerini dene, ×=2⁸=256 sağlıyor.

[@Agegnehu-du8fs]

It's so complicated way to solve 😢

[@user-yr6fg3oe4x]

What is the 3rd solution? It must be a small negative number

[@allozovsky]

Yeah, it's a solution to (1/2)⁻ˣ = (−x)³²

[@alejomdp]

Approximately -0.979.

[@allozovsky]

@@alejomdp How did you get it? Numerically or evaluated via some tool?

[@davidcliftongm8565]

X=0 ?? If 0^x=1 or is that for 0!

[@shriharimutalik1077]

1.022395

[@user-ec5ip3vp2r]

256

[@KyriZee]

there is a faster and more accurate way: x > 0, obvious x cannot be non integer rational number because then LHS = irrational and RHS = rational. So, x > 1 and x is a natural number. Let x be 2 to the power of k. k > 0. So, 2 to the k = 32k. So, 2 to the (k-5) = k. Obvious solution: k = 8, so x = 256. Draw graphs for y = k and y = 2 to the (k-5). For k>8 they cannot meet again (show with differentiation). For 1

[@allozovsky]

@@KyriZee > x > 0, obvious Why is it obvious? Both 2ˣ and x³² are defined for all x ∈ ℝ. > x cannot be non integer rational number But it can be irrational. > I have no idea how to solve for irrational. Using the Lambert W function, obviously. But the solution for an integer x is indeed accurate and neat.

[@evabeyza]

Güzel yöntemmiş ama çok uğraştıcı

[@Bottle134]

How did i get 1

[@entryfragger4478]

Bring log...

[@anderplayerk4222]

x = 0 😎

[@gabrielrocha1557]

ㅉㅊ0/글쿤 ㆍ 3 ⚂ ⛯ ⚅ 2⃣

[@soshakobyan3123]

This is not complete solution, so there are two more solutions, so three in total !!!

[@allozovsky]

There are three real solutions in total: one integer solution and two non-integer solutions (a positive and a negative around 1 by absolute value).

[@soshakobyan3123]

@@allozovsky I know, just I wrote incorrect. I fixed it.

[@user-fr3xf9gj7p]

А если x

[@chrisrardin5043]

😅x=1

[@bobjazz2000]

2 does not equal 1

[@sum6004]

x=0

[@trainer34]

AI Resolving: To solve the equation \( 2^x = X^{32} \) for \( X \) given a specific value of \( x \), follow these steps: 1. Take the natural logarithm (ln) of both sides of the equation: \[ \ln(2^x) = \ln(X^{32}) \] 2. Use the logarithmic identity \( \ln(a^b) = b \ln(a) \): \[ x \ln(2) = 32 \ln(X) \] 3. Solve for \( \ln(X) \) by dividing both sides of the equation by \( 32 \ln(2) \): \[ \ln(X) = \frac{x \ln(2)}{32} \] 4. Exponentiate both sides with the base \( e \) (Euler's number) to solve for \( X \): \[ X = e^{\frac{x \ln(2)}{32}} \] Now, substitute the value of \( x \) into this formula to find the corresponding value of \( X \). For example, if \( x = 4 \): \[ X = e^{\frac{4 \ln(2)}{32}} \approx e^{\frac{\ln(2)}{8}} \approx 1.0905 \] This method allows you to calculate \( X \) for a given value of \( x \). Therefore, for ( x = 4 ), the value of ( X ) is approximately ( 1.0905 ).

[@Karacasah]

Kadın orijinal alman aksanına sahip değil belli ki yabancı uyruklu başka ülkeden gelip almancayı öğrenmiş ...

[@WBKOD.]

X =1 😂

[@semraalbayrak8873]

Çözümü çok uzattınız

[@user-el3yn5gh4p]

😢где тебе это пригодилось в жизни ❓😳

[@Uncle_johanxrz]

Perhaps a normal person wouldn't understand it or need it in real life but the system he surrounded by, use this problem .In order to the world to progress and move forward math is indeed needed

[@crat82]

я надеюсь то, что тебя зачем-то научили писать тебе пригодилось не только в том, чтобы писать тупые комментарии в ютюбах? 😏

[@user-el3yn5gh4p]

@@crat82 мой маленький друг, х+у=5. Найди х и у. И ведь я уверен что ты Найдёшь ❗и зп у тебя станет больше.

[@lunaticlad2022]

​@@user-el3yn5gh4pчто ты тут забыл и если задаёшь такие вопросы))

[@brunomarques1412]

Good process. But it is so annoying how u pronouce Two I had to mute

[@qhawe_tricks]

🤣🤣🤣

[@user-rr2gu9ht2n]

That's because your mom is a whore. It's not his fault.

[@evaristomumba1451]

You ungrateful parasite man's using a language that's not he's

[@LeviAckerman-kt9bt]

Powaa

[@wolfgaming6114]

Nothing is depend on language

[@asianbuilder3344]

Most boring & illogical subject

[@abdullahsuna8400]

X=256