Germany - Math Olympiad Exponential Problem

Рет қаралды 390,579

Күн бұрын

Пікірлер: 222

@dmitrymelnik8296 7 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 5 ай бұрын

Can you explain this more. I don't understand

@cave2song 5 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 4 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.

@alejomdp 7 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 6 ай бұрын

can´t be solved analytically

@albedogray 6 ай бұрын

@@PIANOJOE7why?

@albedogray 6 ай бұрын

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

@adrianlautenschlaeger8578 5 ай бұрын

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

@PIANOJOE7 5 ай бұрын

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

@DakshSood-pv3pl 6 ай бұрын

Very good-question and explanation

@BZKnowHow 6 ай бұрын

DIFFICULT Question but you explain very easily

@yapadek3098 8 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 ?

@ああ-s5m8m 7 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 7 ай бұрын

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

@ああ-s5m8m 7 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 7 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 7 ай бұрын

@@ayugaming3047 Mental arithmetic

@ayugaming3047 7 ай бұрын

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

@maths_withvarsha 3 ай бұрын

Simply awesome 👌. I liked. Nd subscribed too.

@pianoplayer123able 8 ай бұрын

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

@allozovsky 8 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 7 ай бұрын

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

@allozovsky 7 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 7 ай бұрын

@@allozovsky yeah, my bad.

@allozovsky 7 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.

@AtinafAhmad Ай бұрын

I like your way is so amazing

@oliverdauphin236 8 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 7 ай бұрын

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

@maibeez 3 ай бұрын

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

@attica7980 4 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.

@ivangorin1254 7 ай бұрын

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

@antonsemenov5155 Ай бұрын

Иван, каким образом были найдены два других корня (помимо 256)? При проверке корня 1,022393 число сходится только до пятого разряда после точки/запятой, далее отличается

@ivangorin1254 Ай бұрын

Привет Антон! Я применил функцию Ламберта W(ze^z)=z 2^x=x^32 Извлекаем корень из 32 x=+/- 2^(x/32) a=e^(Lna) x=+/-e^[Ln(2^(x/32))]=+/-e^[(xLn2)/32] xe^[(-xLn2)/32]=+/-1 Домножим обе части на -(Ln2)/32 [(-xLn2)/32]e^[(-xLn2)/32]=+/-(Ln2)/32 Возьмём функцию Ламберта от обеих частей W{[(-x(Ln2)/32]e^[(-xLn2)/32]}= W[+/-(Ln2)/32] -x(Ln2)/32=W[+/-(Ln2)/32] x=-32W[+/-(Ln2)/32]/Ln2 При помощи калькулятора для функции Ламберта находим W[+(Ln2)/32]=0,02120634 Ограничимся точностью в 7 значащих цифр x1=-(0,02120634 Ln2)/32=/0,9790196 W[-(Ln2)/32]=-0,02214590 x2=-(-0,02214590 Ln2)/32=1,02239294 Второе значение W[-(Ln2)/32]=-5,545177 x3=-(-5,545177 Ln2)/32=256 Последний корень можно получить точно. Преобразуем выражение -(Ln2)/32 к виду ze^z -(Ln2)/32=-(2^3Ln2)/2^3*32=-(8Ln2)/2^8=-(8Ln2)/e^(8Ln2)=-(8Ln2)*e^(-8Ln2) W[-(8Ln2)*e^(-8Ln2)]=-8Ln2 x3=-(-8Ln2)*32/Ln2=256

@antonsemenov5155 Ай бұрын

@@ivangorin1254 Иван, привет! Благодарю за разъяснение! Теперь понял :)

@jeffersonaraujoelcristiano 3 ай бұрын

It's really otherworldly! Fantastic!

@jkr9224 7 ай бұрын

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

@huzurislamda561 Ай бұрын

Harika(perfect)👏🏻👏🏻👏🏻

@learncommunolizer Ай бұрын

Thank you very much 😊 ☺️ 🙏

@PacoMoyajaquemate 8 ай бұрын

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

@SVC_Editz 7 ай бұрын

Also 1.0223

@sudhangshubhattacharya4991 6 ай бұрын

Very nicely proceeding towards the target

@huseyinaydogan6928 7 ай бұрын

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

@albedogray 6 ай бұрын

why infinite?

@TirthParmar-cw4dc 5 ай бұрын

No bro only one 256 satisfied this equation

@adrianlautenschlaeger8578 5 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 4 ай бұрын

@payoo_2674 2 ай бұрын

@@TirthParmar-cw4dc 3 real solutions: x₁ = 1.0223929402057803206527516798494005683768365119132864517728278977... x₂ = 256 x₃ = -0.979016934957784612322582550011650068748090048886011676265377083... and many complex solutions

@julianbruns7459 7 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 5 ай бұрын

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

@MisiTusi 3 ай бұрын

She was beautiful❤❤❤❤

@oliverdauphin236 8 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

@maths-168 5 ай бұрын

Good excercise.

@ВалентинаКазакова-м3п 7 ай бұрын

Интересное задание и его решение!

@hoahocphothongmoi2212 6 ай бұрын

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

@سمبوساقيمز 7 ай бұрын

just use the logarithm equation

@InnocentMsomi-ti4lx 3 ай бұрын

It is impossible in this case

@nefando7777 7 ай бұрын

Me encanta cómo dice "two".

@Rao_adib 6 ай бұрын

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

@Eleuthero5 7 ай бұрын

Wow that was a very entertaining proof!!

@learncommunolizer 7 ай бұрын

Thank you very much!

@xgx899 7 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 4 ай бұрын

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

@darthmetatron 4 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 3 ай бұрын

Make your video. I wanna see your solutions

@Stay_EU_Independence 7 ай бұрын

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

@dinlendiricidrtv 6 ай бұрын

Because 256

@guilhermenoronha1978 5 ай бұрын

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

@OlympiadMentor 6 ай бұрын

Tricky Question but you explain very well ❤.

@ozodausmonova18 5 ай бұрын

It was a lengthy solution but wonderful

@learncommunolizer 5 ай бұрын

Your welcome

@richardleveson6467 8 ай бұрын

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

@muhammadarshadh633 7 ай бұрын

That's what she said

@kanguru_ 2 ай 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.

@claudebalzano7031 7 ай бұрын

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

@FFATITUITEBOYOFFCIAL 7 ай бұрын

Very good solution

@gregsti 8 ай бұрын

Très intéressant !

@learncommunolizer 7 ай бұрын

Thank you very much!

@VoiceTuned 6 ай бұрын

Wow very nice

@AKRiitMedPhysicsGuru 6 ай бұрын

Nice bro🎉

@payoo_2674 2 ай бұрын

Use the Lambert W function W(■*e^■) = ■ 2^x = x^32 ln(2^x) = ln(x^32) x*ln(2) = 32*ln|x| ===> two cases 1st case: x > 0 x*ln(2) = 32*ln(x) ln(x)*x^(-1) = ln(2)/32 ln(x)*e^ln(x^(-1)) = ln(2)/32 ln(x)*e^(-ln(x)) = ln(2)/32 -ln(x)*e^(-ln(x)) = -ln(2)/32 W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32) -ln(x) = W(-ln(2)/32) ln(x) = -W(-ln(2)/32) x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977... in WolframAlpha: e^(-productlog(0,-ln(2)/32)) x₂ = e^(-W₋₁(-ln(2)/32)) = 256 in WolframAlpha: e^(-productlog(-1,-ln(2)/32)) 2nd case: x < 0 x*ln(2) = 32*ln(-x) ln(-x)*x^(-1) = ln(2)/32 -ln(-x)*x^(-1) = -ln(2)/32 ln(-x)*(-x)^(-1) = -ln(2)/32 ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32 ln(-x)*e^(-ln(-x)) = -ln(2)/32 -ln(-x)*e^(-ln(-x)) = ln(2)/32 W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32) -ln(-x) = W(ln(2)/32) ln(-x) = -W(ln(2)/32) -x = e^(-W(ln(2)/32)) x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083... in WolframAlpha: -e^(-productlog(0,ln(2)/32))

@Afriexfull 2 ай бұрын

YOU TALK WELL SIR,

@HiệpNguyễnTấn-v1r Ай бұрын

GOOD IDEA

@salehabasov7074 8 ай бұрын

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

@yaasiinbekham6935 3 ай бұрын

Thank you sir❤

@WenjieGu 7 ай бұрын

Très intéressant

@learncommunolizer 7 ай бұрын

Thank you very much !

@RijuPatra-t3p Күн бұрын

The process is good for written tests but it's too long for compitive exams.

@knotwilg3596 4 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).

@BrianDroid-g2d 8 ай бұрын

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

@allozovsky 8 ай бұрын

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

@alejomdp 7 ай бұрын

Approximately -0.979.

@allozovsky 7 ай бұрын

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

@guilhermenoronha1978 5 ай бұрын

@@allozovsky wolfram alpha, it's a calculator

@nahideipekcioglu7552 7 ай бұрын

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

@ЕленаЛарина-я8я 5 ай бұрын

Сочетание логики и метода научного тыка

@noname-ed2un 5 ай бұрын

Can i use the lambert W function

@nazmurrahmantamim6014 4 ай бұрын

W Lambert is useable for this

@payoo_2674 2 ай бұрын

@@nazmurrahmantamim6014 YES! 2^x = x^32 ln(2^x) = ln(x^32) x*ln(2) = 32*ln|x| ===> two cases 1st case: x > 0 x*ln(2) = 32*ln(x) ln(x)*x^(-1) = ln(2)/32 ln(x)*e^ln(x^(-1)) = ln(2)/32 ln(x)*e^(-ln(x)) = ln(2)/32 -ln(x)*e^(-ln(x)) = -ln(2)/32 W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32) -ln(x) = W(-ln(2)/32) ln(x) = -W(-ln(2)/32) x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977... x₂ = e^(-W₋₁(-ln(2)/32)) = 256 2nd case: x < 0 x*ln(2) = 32*ln(-x) ln(-x)*x^(-1) = ln(2)/32 -ln(-x)*x^(-1) = -ln(2)/32 ln(-x)*(-x)^(-1) = -ln(2)/32 ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32 ln(-x)*e^(-ln(-x)) = -ln(2)/32 -ln(-x)*e^(-ln(-x)) = ln(2)/32 W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32) -ln(-x) = W(ln(2)/32) ln(-x) = -W(ln(2)/32) -x = e^(-W(ln(2)/32)) x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...

@davez8816 3 ай 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??

@Mralipol 6 ай бұрын

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

@Austin101123 7 ай бұрын

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

@Bishirii-fu9wf 2 ай бұрын

Thonk you ❤❤❤❤❤

@himadrikhanra7463 7 ай бұрын

(1)^x/32=2^32 ×x...so x= 1/2^32....

@pujarian 7 ай бұрын

Just take log both sides

@TediTeda 4 ай бұрын

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

@VolodymyrBoyka Ай бұрын

Log without calculator is not possible in math Olympiad that's where IQ comes in

@ronaldnoll3247 5 ай бұрын

excellent

@learncommunolizer 5 ай бұрын

Thank you! Cheers!

@totalhere Ай бұрын

Wow Chat GPT can not answer. but you can.

@evabeyza 7 ай бұрын

Güzel yöntemmiş ama çok uğraştıcı

@TumeloPercy-z3s 3 ай бұрын

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

@DacadYare-k2t 7 ай бұрын

Thanks

@N0B0DY_SP3C14L Ай бұрын

Whoa hold up. 2*1/32= 2/32 or 1/16. When multiplying a fraction by an integer, the divisor stays the same or simplifies. You are only performing the multiplication on the numerator. 1/16=1/x. X=16. Why make this more difficult?

@永辉万-i3n 5 ай бұрын

are u from India?

@Phạm_Quang_Minh100 2 ай бұрын

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

@namantiwari4529 7 ай бұрын

Finnally bro did it

@sureshgattu9243 6 ай бұрын

You can solve using the concept of logarithms

@liquidzliquidz3557 5 ай бұрын

How I try to solve

@SouvereineMaman 8 ай бұрын

Not Bad like methode

@learncommunolizer 8 ай бұрын

Thank you very much!!!

@Joe-b4c7n 8 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 🇩🇴⚔️☕

@darthmetatron 4 ай бұрын

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

@nirmalyadatta3523 4 ай бұрын

👍🙏

@darthmetatron 4 ай бұрын

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

@Agegnehu-du8fs 7 ай бұрын

It's so complicated way to solve 😢

@MisiTusi 3 ай бұрын

She is a beautiful ❤❤❤❤

@davidcliftongm8565 7 ай бұрын

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

@alfredteichmann135 7 ай бұрын

Take two sheets of paper!

@mastanvali255 4 ай бұрын

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

@worldorthoorthopaedicsurge6147 2 ай бұрын

At 71, x is 2 or -2

@kennycovenas3598 6 ай бұрын

Muy complejo el procedimiento

@rslua01 7 ай бұрын

@edfess5424 Ай бұрын

the answer is 32x32=1024 2x2=4 1024/4=256

@АндрейЛюбавин-э4щ 8 ай бұрын

256

@KyriZee 8 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 7 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.

@ivangorin1254 Ай бұрын

Я применил функцию Ламберта W(ze^z)=z 2^x=x^32 Извлекаем корень из 32 x=+/- 2^(x/32) a=e^(Lna) x=+/-e^[Ln(2^(x/32))]=+/-e^[(xLn2)/32] xe^[(-xLn2)/32]=+/-1 Домножим обе части на -(Ln2)/32 [(-xLn2)/32]e^[(-xLn2)/32]=+/-(Ln2)/32 Возьмём функцию Ламберта от обеих частей W{[(-x(Ln2)/32]e^[(-xLn2)/32]}= W[+/-(Ln2)/32] -x(Ln2)/32=W[+/-(Ln2)/32] x=-32W[+/-(Ln2)/32]/Ln2 При помощи калькулятора для функции Ламберта находим W[+(Ln2)/32]=0,02120634 Ограничимся точностью в 7 значащих цифр x1=-(0,02120634 Ln2)/32=/0,9790196 W[-(Ln2)/32]=-0,02214590 x2=-(-0,02214590 Ln2)/32=1,02239294 Второе значение W[-(Ln2)/32]=-5,545177 x3=-(-5,545177 Ln2)/32=256 Последний корень можно получить точно. Преобразуем выражение -(Ln2)/32 к виду ze^z -(Ln2)/32=-(2^3Ln2)/2^3*32=-(8Ln2)/2^8=-(8Ln2)/e^(8Ln2)=-(8Ln2)*e^(-8Ln2) W[-(8Ln2)*e^(-8Ln2)]=-8Ln2 x3=-(-8Ln2)*32/Ln2=256

@vishnujatav6329 6 ай бұрын

Tricky

@anderplayerk4222 7 ай бұрын

x = 0 😎

@sevincsultanova7649 4 ай бұрын

👍

@learncommunolizer 4 ай бұрын

👍👍👍

@miracbarsapaydn650 8 ай бұрын

X = 256

@AshetuNegasa 4 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

@ЕленаЛиненко-к7у 4 ай бұрын

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

@payoo_2674 2 ай бұрын

Или использoвaть функцию Ламберта W 2^x = x^32 ln(2^x) = ln(x^32) x*ln(2) = 32*ln|x| ===> два случая 1°: x > 0 x*ln(2) = 32*ln(x) ln(x)*x^(-1) = ln(2)/32 ln(x)*e^ln(x^(-1)) = ln(2)/32 ln(x)*e^(-ln(x)) = ln(2)/32 -ln(x)*e^(-ln(x)) = -ln(2)/32 W(-ln(x)*e^(-ln(x))) = W(-ln(2)/32) -ln(x) = W(-ln(2)/32) ln(x) = -W(-ln(2)/32) x = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> два корня x₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977... x₂ = e^(-W₋₁(-ln(2)/32)) = 256 2°: x < 0 x*ln(2) = 32*ln(-x) ln(-x)*x^(-1) = ln(2)/32 -ln(-x)*x^(-1) = -ln(2)/32 ln(-x)*(-x)^(-1) = -ln(2)/32 ln(-x)*e^ln((-x)^(-1)) = -ln(2)/32 ln(-x)*e^(-ln(-x)) = -ln(2)/32 -ln(-x)*e^(-ln(-x)) = ln(2)/32 W(-ln(-x)*e^(-ln(-x))) = W(ln(2)/32) -ln(-x) = W(ln(2)/32) ln(-x) = -W(ln(2)/32) -x = e^(-W(ln(2)/32)) x = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> один корень x₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083...