Germany - Math Olympiad Exponential Problem

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Күн бұрын

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Пікірлер: 199

@dmitrymelnik8296 5 ай бұрын

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.

@noname-ed2un 2 ай бұрын

Can you explain this more. I don't understand

@cave2song 2 ай бұрын

2^x = x^32 ln2 2^x = ln2 x^32 x ln2 2 = 32 ln2 x x = 32 ln2 x x / (ln2 x) = 32 x x / (ln2 x) < 0 -> x must larger than 1 x = 256 -> 256 / (ln2 256) = 32

@knotwilg3596 2 ай бұрын

@@noname-ed2un He starts from the assumption that X will be a power of 2. This is a good start for finding a solution, since it reduces the complexity of the problem, because both sides are now powers of two. Set x = 2^m 2^(2^m) = (2^m) ^32 2^(2^m) = 2^32m ------ power of power --> multiply exponents 2^m = 32m ---- equal powers --> equal exponents Now this is still not trivial but left side is a power of 2 and right side has 32, which is also a power of 2. Hence m itself must be a power of 2. Set m = 2^k 2^(2^k) = 32 * 2^k 2^(2^k) = 2^5 * 2^k 2^(2^k) = 2^(5+k) 2^k = 5+k Now we're down to small numbers and we can do a quick check for k = 1, 2, 3 ... to find 2^3 = 8 = 5+3 So m = 8 and x = 2^8 = 256 This was a bit lengthy but hopefully helpful.

@ivangorin1254 4 ай бұрын

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

@PacoMoyajaquemate 5 ай бұрын

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

@SVC_Editz 5 ай бұрын

Also 1.0223

@pianoplayer123able 5 ай бұрын

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

@allozovsky 5 ай бұрын

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 5 ай бұрын

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

@allozovsky 5 ай бұрын

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 5 ай бұрын

@@allozovsky yeah, my bad.

@allozovsky 5 ай бұрын

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.

@سمبوساقيمز 5 ай бұрын

just use the logarithm equation

@InnocentMsomi-ti4lx Ай бұрын

It is impossible in this case

@evabeyza 4 ай бұрын

Güzel yöntemmiş ama çok uğraştıcı

@ああ-s5m8m 4 ай бұрын

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 4 ай бұрын

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

@ああ-s5m8m 4 ай бұрын

@@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 4 ай бұрын

@@ああ-s5m8m your English is fine. So, what calculations did you use after the step 2^x = 32t? Did you just guess?

@ああ-s5m8m 4 ай бұрын

@@ayugaming3047 Mental arithmetic

@ayugaming3047 4 ай бұрын

@@ああ-s5m8m can you introduce that to me?

@BZKnowHow 3 ай бұрын

DIFFICULT Question but you explain very easily

@Mralipol 4 ай бұрын

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

@Austin101123 4 ай бұрын

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

@knotwilg3596 2 ай бұрын

You spend an enormous amount of time finding the integer solution, while you fail to discuss the non-integer solutions (or complex solutions). Math is not just endless steps of algebra to arrive at prior knowledge. Math is about elegance in finding the solution and about rigor in figuring out if there are more solutions (depending on the solution space, which you need to define first).

@semraalbayrak8873 4 ай бұрын

Çözümü çok uzattınız

@kanguru_ Сағат бұрын

2^x=x^32, so xlog2=32logx, then x/logx=32/log2=2^5/log2=2^6/log(2^2)=2^7/log(2^4)=2^8/log(2^8), so x = 2^8.

@asianbuilder3344 5 ай бұрын

Most boring & illogical subject

@nahideipekcioglu7552 4 ай бұрын

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

@yapadek3098 5 ай бұрын

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

@julianbruns7459 4 ай бұрын

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?

@noname-ed2un 2 ай бұрын

That's why I was thinking of using natural log and the Lambert W function

@MisiTusi 15 күн бұрын

She was beautiful❤❤❤❤

@Joe-b4c7n 5 ай бұрын

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

@davez8816 15 күн бұрын

We have a^2=2^2 leading to 2 solutions which are -2 and 2. What make you believe that the fact we have a^(1/a)=x^(1/x) gives a unique solution a=x??

@alejomdp 4 ай бұрын

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 4 ай бұрын

can´t be solved analytically

@albedogray 4 ай бұрын

@@PIANOJOE7why?

@albedogray 4 ай бұрын

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

@adrianlautenschlaeger8578 2 ай бұрын

@@PIANOJOE7 it is possible! Take a look at the Lambert-W-Function. In Wolframalpha to be used as productlog(x).

@PIANOJOE7 2 ай бұрын

@@adrianlautenschlaeger8578 Well, good hint. This function was unkown to me (or already forgoten 🙂...). So you´re right.

@AshetuNegasa Ай бұрын

But at the end of your working; it doesn't make sense ;how took it( 256 the power of 1over 256 is equal x 1 over x ) how you equated But i appreciate you wow.... Can you say sth on that , what i said

@huseyinaydogan6928 4 ай бұрын

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

@albedogray 4 ай бұрын

why infinite?

@TirthParmar-cw4dc 3 ай бұрын

No bro only one 256 satisfied this equation

@adrianlautenschlaeger8578 2 ай бұрын

@@TirthParmar-cw4dc There are infinite complex solutions. Complex numbers are defined by their absolute value and their argument, which is just the angle in the complex plane. And you can add 2πz (z∈Z) to the argument, then you get the same complex number again with a different angle. (2π = 1 full circle) Example: -1 = e^(iπ) = e^(i*(π+2π) = e^(3iπ) = e^(5iπ) = e^(7iπ) = ...

@mikhailhop848 Ай бұрын

@darthmetatron Ай бұрын

looks like a binary problem, x guesses would be 1, 2, 8, 16, 32.....256! No algeabras here!

@centerofvaluableskills 2 ай бұрын

Basic concept of differential equation: kzbin.info/www/bejne/i6qzo6SJms2Xq6csi=01ChXOFTvIS5j-sB

@marceliusmartirosianas6104 26 күн бұрын

|||| 2^x=x^32|||||| X=16 ||| ACADEMIC UNIVERSITA della Columbia MARCELIUS MARTIROSIANAS

@marceliusmartirosianas6104 26 күн бұрын

Kadangi :|||||||||| X^32 =32 tada:||||||||||| 2^x= 32 x= 16 |||| ACADEMIC MARCELIUSD MARTIROSIANAS NOBEL PRIZE Fermatian Prize FELDS MEDALIST

@永辉万-i3n 3 ай бұрын

are u from India?

@Rao_adib 4 ай бұрын

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

@Agegnehu-du8fs 4 ай бұрын

It's so complicated way to solve 😢

@maths_withvarsha 18 күн бұрын

Simply awesome 👌. I liked. Nd subscribed too.

@sum6004 4 ай бұрын

x=0

@ЕленаЛиненко-к7у Ай бұрын

Это не решение, а подбор только одного из трех корней. Количество трех корней можно увидеть графически.

@davidcliftongm8565 4 ай бұрын

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

@kennycovenas3598 4 ай бұрын

Muy complejo el procedimiento

@mastanvali255 Ай бұрын

By applying logarithm, (Logx/Log3) = x/9. Next step??

@darthmetatron Ай бұрын

It is just a binary number problem. ((32)^2)/(2^2) = 256

@TumeloPercy-z3s Ай бұрын

Can the Lambert W-Function help us to find the solution to this equation?

@DakshSood-pv3pl 3 ай бұрын

Very good-question and explanation

@Bottle134 3 ай бұрын

How did i get 1

@nefando7777 5 ай бұрын

Me encanta cómo dice "two".

@alfredteichmann135 5 ай бұрын

Take two sheets of paper!

@ЕленаЛарина-я8я 2 ай бұрын

Сочетание логики и метода научного тыка

@mikaelvilhjalmsson6941 8 күн бұрын

How do I solve this: 2^x + 2^(x+1)? Why is it 3* 2^x ? I'm blind, I just can't see it? =D

@mikaelvilhjalmsson6941 8 күн бұрын

How do I factor 2^x + 2^(x+1) ?

@noname-ed2un 2 ай бұрын

Can i use the lambert W function

@nazmurrahmantamim6014 2 ай бұрын

W Lambert is useable for this

@oliverdauphin236 5 ай бұрын

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

@sureshgattu9243 3 ай бұрын

You can solve using the concept of logarithms

@liquidzliquidz3557 2 ай бұрын

How I try to solve

@Stay_EU_Independence 5 ай бұрын

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

@dinlendiricidrtv 4 ай бұрын

Because 256

@guilhermenoronha1978 2 ай бұрын

yeah aprox. 1.022, for that you have to use the lambert function (w). Also another answer would be aprox. -0.979

@hameratahir Ай бұрын

why are they called Arabic numbers?

@claudebalzano7031 5 ай бұрын

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

@gabrielrocha1557 4 ай бұрын

ㅉㅊ0/글쿤 ㆍ 3 ⚂ ⛯ ⚅ 2⃣

@xgx899 4 ай бұрын

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".

@darthmetatron Ай бұрын

Three guess' I got it. I like to solve things by experience and guesses too.

@darthmetatron Ай бұрын

Just looked back at the problem and I understand You, there are infinite solutions if you do not have use Integers, they did not specify it so...

@nahuelastor7522 27 күн бұрын

Make your video. I wanna see your solutions

@shriharimutalik1077 4 ай бұрын

1.022395

@pspprabhat Ай бұрын

×^+/×=2^1/32 =2^2/2^6=4^1/64 =4^2/128=16^128 =16^2/256=256^1/256 =>×=256

@anderplayerk4222 4 ай бұрын

x = 0 😎

@antoniomantilla7080 2 күн бұрын

too much work and get lost try to remember

@entryfragger4478 4 ай бұрын

Bring log...

@oliverdauphin236 5 ай бұрын

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 5 ай бұрын

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

@maibeez Ай бұрын

Pourquoi avez vous mis x=2^n alors que vous ne connaissez pas sa valeur exacte? Expliquez moi s'il vous plaît

@pujarian 5 ай бұрын

Just take log both sides

@TediTeda Ай бұрын

Hello dear I am from Ethiopia and I don't understand logarithm so do u know any good channel to help me

@alexsandrodasilva3699 Ай бұрын

A pior explicação de toda a minha vida

@noname-ed2un 2 ай бұрын

2^x = x^32 ln2^x = ln x^32 Xln2=32ln x ln2/32 = lnx/x ln 2/32 ln x/ e^lnx ln 2/32=lnx *e^-lnx - ln2/32 = -lnx *e^-lnx W(-ln2/32) = W( -lnx * e^ -lnx) W(-ln2/32) = -lnx -W(-ln2/32) = x Put the function in wolframalpha and you get all the solutions

@soshakobyan3123 5 ай бұрын

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

@allozovsky 5 ай бұрын

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 5 ай бұрын

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

@miracbarsapaydn650 5 ай бұрын

X = 256

@hoahocphothongmoi2212 4 ай бұрын

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

@ahmadibrahim4075 10 күн бұрын

Just take log of both side

@maths-168 2 ай бұрын

Good excercise.

@Phạm_Quang_Minh100 8 күн бұрын

Tôi thấy phương pháp logarit là ngắn gọn, và thuyết phục nhất

@vishnujatav6329 4 ай бұрын

Tricky

@himadrikhanra7463 5 ай бұрын

(1)^x/32=2^32 ×x...so x= 1/2^32....

@ОлегМехов-в6р 4 ай бұрын

А если x

@salehabasov7074 5 ай бұрын

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

@MisiTusi 15 күн бұрын

She is a beautiful ❤❤❤❤

@ConvivialWorldTravel 4 ай бұрын

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

@ЕленаЛиненко-к7у Ай бұрын

В этом уравнении тоже три решения. Постройте графики. Но выразить их через трансцендентные функции сложно. Решение либо графическое, либо приближенными численными методами.

@yaasiinbekham6935 12 күн бұрын

Thank you sir❤

@attica7980 Ай бұрын

In addition to the solution x=256, the equation also has a real solution between -1 and 0. With a problem like this, it would be nice to be specific whether to find all real solutions, or to find only solutions that are positive integers.

@jeffersonaraujoelcristiano 14 күн бұрын

It's really otherworldly! Fantastic!

@BrianDroid-g2d 5 ай бұрын

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

@allozovsky 5 ай бұрын

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

@alejomdp 4 ай бұрын

Approximately -0.979.

@allozovsky 4 ай бұрын

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

@guilhermenoronha1978 2 ай бұрын

@@allozovsky wolfram alpha, it's a calculator

@chrisrardin5043 5 ай бұрын

😅x=1

@bobjazz2000 4 ай бұрын

2 does not equal 1

@DacadYare-k2t 4 ай бұрын

Thanks

@namantiwari4529 4 ай бұрын

Finnally bro did it

@richardleveson6467 5 ай бұрын

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

@muhammadarshadh633 4 ай бұрын

That's what she said

@ВалентинаКазакова-м3п 4 ай бұрын

Интересное задание и его решение!

@FFATITUITEBOYOFFCIAL 4 ай бұрын

Very good solution

@АндрейИванов-д2з 7 күн бұрын

2*32

@AnkitaKumari-hz3ms Ай бұрын

X=256

@GeorgetaBobar 2 ай бұрын

X= 256

@bryananthony2192 2 ай бұрын

x = 8

@AKRiitMedPhysicsGuru 4 ай бұрын

Nice bro🎉

@اسدرضایی-و6ي 2 ай бұрын

😂😂😂😂👎

@VapreccolarColar9 27 күн бұрын

@АндрейЛюбавин-э4щ 5 ай бұрын

256

@KyriZee 5 ай бұрын

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 5 ай бұрын

@@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.