Math Olympiad | How to solve for X in this problem ?

[@juancarlosnadermora716]

One elevated to any real number is one. So it can’t be equal to two. Therefore you have to look for a complex resolution.

[@yiutungwong315]

1^All Real Number = 1

[@Jan_D-vm5rk]

Hello, You use the formula ln(a^b)=b*ln(a) but this is generally not true for complex numbers. To prove this let us consider ln(1). We know that ln(1)=0. But 1 can also be written as e^(2πi). So if the formula were correct we could write ln(1)=ln[e^(2πi)]= 2πi*ln(e)=2πi. Because 1 can also be written as e^(4πi) we could prove in a similar way that ln(1)=4πi. So ln(1) would be both 0, 2πi, 4πi and so on. Of course this can not be the case. The formula (e^a)^b=e^(a*b) is generally not true for complex numbers either. Suppose it were true then let us consider 1^i. Because 1=e^0 we could write 1^i=(e^0)^i=e^(0*i)=e^0=1 But also 1=e^(2πi) so this would give 1^i=[e^(2πi)]^i=e^(2πi*i)=e^(-2π). So 1^i would be both 1 and e^(-2π). Of course this can not be true. In order to know what 1^b is for an arbitrary complex number b we need to know its definition. For complex numbers a and b, a^b is defined as e^[b*ln(a)] So in the case 1^b, 1^b=e^[b*ln(1)]=e^(b*0)=e^0=1 So for all complex b, 1^b=1. We have not defined so far what e^z is for an arbitrary complex z. When z is written as z=a+ib, where a and b are real, e^z can be defined in such a way that: e^z=e^(a+ib)=(e^a)*[cos(b)+isin(b)] For all complex z and w it can be proven that (e^z)*(e^w)=e^(z+w) We still have to define ln(z). Every complex z which is not 0 can be written in the form z= r*[cos( φ)+isin(φ)] where r and φ are real and r>0 and -π

[@dllord]

For sin and cos 0,2π,4π,... it is the same sin(0)=sin(2π)=sin(4π) And same with cos

[@Jan_D-vm5rk]

@@dllord Yes, and that is not a problem and allowed. But it would be a problem and not allowed when sin(0)=0 and also sin(0)=1. A better comparison is that it is allowed that 2^2=4 and (-2)^2=4 but √4=2 and it is not allowed that also √4= -2 because a function has to be uniquely defined.

[@salvatorecosta875]

No! For complex numbers ln1 =2πki

[@beardslay]

@@Jan_D-vm5rkfunctions do not have to be uniquely defined. this is a property of some functions but not all. some functions are single-valued, other functions are multuvalued. sqrt and arccos are multivalued on the real line. log is multivalued in the complex plane.

[@Jan_D-vm5rk]

@@beardslay Okay, I understand what you mean. This is not a common view. Go for example to the site: wolframalpha , which is very good at mathematics. Type x^2=4 and you get the answers 2 and -2. Type sqrt(4) and you get the only answer 2. Type sin(x)=0 and you get the answer x=n π Type arcsin(0) and you get the only answer 0. Type e^x=i and you get x=1/2 i (π + 4 n π) Type log(i) and you get the only answer (i π)/2. This is in accordance with the definition given by Encyclopaedia Britannica: The modern definition of function was first given in 1837 by the German mathematician Peter Dirichlet: If a variable y is so related to a variable x that whenever a numerical value is assigned to x, there is a rule according to which a unique value of y is determined, then y is said to be a function of the independent variable x. This relationship is commonly symbolized as y = f(x)-which is said “f of x”-and y and x are related such that for every x, there is a unique value of y. That is, f(x) can not have more than one value for the same x.

[@thadeu77]

Congratulations!!! I'm Brazilian and, in addition to the very well demonstrated exercise, i congratulate you on these spectacular lyrics...

[@richardleveson6467]

It's interesting - that which is impossible in the real world can happen in the imaginary one. In other words you can find powers of one that are not equal to one IF you 'pretend' that square roots of negative numbers exist! Note however that getting an infinite number of imaginary values for x is a little unsettling! Thanks for this clever video!

[@abtinnavid6903]

Complex number system is a number system. The word imaginary was used by René Descartes as he did not agree with the concept. But the name stuck! The word lateral is a better name.

[@abtinnavid6903]

Without complex numbers, various mathematical and scientific applications would be limited. Complex numbers is as real ans the Real number system and we don’t need to pretend. They play a vital role in fields such as electrical engineering, quantum mechanics, signal processing, and fluid dynamics. Removing complex numbers would restrict our ability to represent and analyse phenomena with oscillatory behavior or systems involving lateral quantities.

[@abtinnavid6903]

By the way the solution is is incorrect and hence, rubbish.

[@rubendariomartin728]

Hola, tengo un manejo muy pobre del Inglés, pero los signos matemáticos se entienden bien, una elegante resolución, me parece. Gracias por compartir a la persona que resolvió y gracias a KZbin.

[@dugasdugas794]

Math Olympiad | How to solve for X in this problem ?👍👍👍

[@mhtsakos]

There is something i don't understand.. if you replace the solution in the original equation you get 1^(a*i)=2 in which a equals to -ln(2)/2κπ so 1^(a*i)=(1^a)^i=1^i Doesn't seem right

[@dlmperplex337]

The mistake is that he's cancelling zeros, albeit in the exponent

[@salvatorecosta875]

that's wright! if t=1^(ai) ln(t)=ln(1^(ai))=ai*ln1=ai*2πki=-2πka =-2πk*(-ln(2)/2κπ)=ln2 then t=1^(ai)=2. The secret is remembar that ln1 is not zero but ln1 is not zero but in general ln1=2πki.

[@salvatorecosta875]

Perhaps it is useful to observe that 1= e^(2πki) therefore ln1=2πki (in general ln1 is not zero). equation 1^x=2 becomes x*ln1=ln2 then x=(ln2)/2πki

[@Jdelarua2023]

But i dont understand. Number 1 of the power of any number is always 1. Even if number 1 of the power of a complex number, doesn't has sense because 1 plus itself in complex way can't be possible, if the base is a real number. Is imposibble to obtain a value for "x" to satisfy the equation, because will never equal to 2. Doesn't it?

[@salvatorecosta875]

@@Jdelarua2023 When working with complex numbers, abandon all obviousness. You only have to refer to the calculation rules and, mainly, to Euler's formulas and the Moivre rule. So 1= e^(2πki) for Euler (in fact cos 2πki+i sen2πki=1+i*0=1). 1^n is not 1 but [e^(2πki)]^n then, for Moivre 1^n=e^(2nπki) For n integer, nk is integer 1^x=e^(2lπi) with l=nk integer then 1^n=1. For n=x ,with x complex , 1^x=[e^(2πki)]^x=e^(2πkxi) with k integer and complex x which can be written as x=a+ib (with real a,b) . therefore 1^x=e^[2πki(a+ib)]= e^[2πkai-2πkb]=[e^(-2πkb)]*[e^(2πkai)] but e^(-2πkb)=ρ is real. Then 1^x=ρ[cos2πka+isin2πka] which is not 1 but a complex number.

[@simonkara4907]

Greetings. A peculiar answer you've provided: 1 is a constant and not a constant at the same time. Clearly, this is an absurdity, that implicates a flaw in the reasoning that led to it. e^ix is a function from the complexe plane to the complexe plane. The "ln" function, however, is the inverse of the real valued function x->e^x that takes real values in. Thus, your use of the ln on x->e^ix, as if it was a real valued function and its inverse, is incorrect.

[@salvatorecosta875]

@@simonkara4907 The rules valid for real numbers are extended to the complex field so that lnx exists both for positive and negative real x and for imaginary or complex x. e.g. lni exists and we have lni=ln(e^i(π/2+2kπ)=i(π/2+2kπ) similarly ln(-1)=lne^i(π+2πk)=i(π+2πk)

[@marcusgloder8755]

Hello everyone, there are a few things about the video that I find strange. On the one hand, the tutor puts in a lot of mathematical effort. All you have to do is solve the equation for x, which gives x = log₁(2), and enter it in this form into any CAS to find out that x is equal to infinity. The second thing I don’t understand is that, despite using advanced mathematical techniques, the tutor obviously divides by zero in at least two places and then thinks the result is zero. This is the case at minute 1:19 and at minute 3:39. Every child knows that you can’t divide by zero. That’s a huge prohibition sign. Anyone who uses advanced techniques like the tutor should actually know this. And he apparently knows it too, as evidenced by something he writes at minute 7:21. But that just makes it all even more mysterious. Best regards Marcus 😎

[@ianeinman1880]

He is not dividing by zero, he's just using a confusing notation to show that the expression above the line simplifies to the expression below the line. cos(0) -> 1 and sin(0) -> 0. He's underlining the expression as he talks, and then writing its numeric value under it.

[@222aint]

beautiful solutions

[@ガアラ-h3h]

This one is actually impossible since any number taken to the power of a complex number must be in the interval inside the circle with radius one of the complex plain

[@cwcarson]

Neat handwriting

[@learncommunolizer]

Thank you very much 😊

[@demetrioalbuquerque5408]

👏👏 show de resolução Mestre!!

[@learncommunolizer]

Thanks so much 🙏❤️🙏

[@kangsungho1752]

very kind explaination!

[@HassanAli-sn3ch]

You're an genius!!!

[@redroach401]

Eulers formula is re^itheta dont forget the radius

[@samuellincoln]

Wouldn't it be straightforward that x = log of 2 in basis 1?

[@math_travel]

I begin my today with this interesting problem. I was wondering what the answer was. because I knew that x could not be real number. you applied the euler's formula. I couldn't expect that. thanks for good morning~~

[@learncommunolizer]

Thank You very much! 👍👍 You are very welcome!

[@charlamps]

Very nice resolution.

[@ГригорийРайцес]

На сколько я помню правило из школы- 1 в любой степени равно 1! Комплексные числа в моё время в школе не изучали, а только в институте! Поэтому, по школьной программе, здесь решений быть не может!!!

[@wolfgang8001]

А о чем ролик? Смотри внимательно и пойми, что он написал, что нет решений в действительных числах

[@ГригорийРайцес]

@@wolfgang8001 поэтому и пишу, что не могу понять для кого ролик! Для школьников? Но комплексные числа, в моё время, не знаю как сейчас, в школе не проходили!

[@mpguedj]

I don't understand. At the beginning we want to get sin0 is equal to 0. so, there are two solutions: sin0 = zero and sin π = 0. Therefore the final formula must be written as follow: x= -i ln2 /kπr

[@manueld848]

Shouldn't we also treat 2 as a complex number? Of course we could continue to infinity if we also treat the 2 of said solution as complex.

[@Bruh-bk6yo]

2 is already a complex number💀 2+0*i

[@manueld848]

@@Bruh-bk6yo Of course; If it were not, we could not treat it as such. But in the same way that he has done with 1, we can put 2 as 2·e^(i·2kπ), whose logarithm is Ln(2) + i·2kπ. Once again we have a 2 which in turn we would treat again as complex, and so on ad infinitum.

[@Bruh-bk6yo]

@@manueld848 your logic is clear and it's correct. e^(i*2πn)=e^(ln2) 2 does not have any imaginary part, therefore, the Arg2 function equals 0, that's why we treat 2 as a "real" number. Not complex.

[@manueld848]

@@Bruh-bk6yo And what is special about number 1 of the exercise? Aren't we in the same case?

[@Bruh-bk6yo]

@@manueld848 the fact that 1 raised to the power of variable x.

[@stathislourantos1194]

I think that the solution starts being incorrect at 5:00 where ln(LHS) is assumed to be equal with ln(RHS). This might be correct in one direction but when solving an equation every step needs to work both ways. In other words ln(LHS)=ln(RHS) does not necessarily mean that LHS=RHS when we are dealing with complex numbers (we agreed that there is no real solution). Things are different with real numbers because ln(x) is a strictly increasing function therefore ln(x)=ln(y) ⟺ x=y. But such an assumption is invalid in the complex plane. Besides, the solution cannot be confirmed. If we substitute x witht he solution provided in the initial equation we will end up with 1^(-i)=2 or 1^i=1/2. And if we solve the equation 1^x=3 and substitute the solution (containing ln(3) this time), we will end up with 1^i=1/3. Similarly 1^i=1/4 and so forth... So since it cannot be that 1/2=1/3=1/4, the method elaborated is rubbish...

[@mrjnutube]

Oh yes, the solution CAN BE CONFIRMED. Just plug in (-iln2/2kπ) for "x", and you'll get e^ln2, which is 2. Make sure everything in the equation (LHS) 1^x is in the Complex realm.

[@skyrider53]

What was that background noise at 1:23?

[@魔色]

Euler: Who is Eurle?😂😂😂😂

[@michaelhartmann1285]

The initial equation was ridiculous on its face. 1 raised to any power is still 1, so there was no point even fooling with it.

[@pelasgeuspelasgeus4634]

Eurler's??? Really now? I would accept your solution if you could graph it. Can you?

[@catBasilio]

Математика - отличный способ заморочить голову на ровном месте.

[@narolenkoaleksandr58]

Изобретаем вечный двигатель😅😅😅!!!???

[@wadikz2585]

Ну, а вдруг... Доказывают же в YT, что Земля плоская

[@danman6669]

It's actually Euler's Formula, not Eurler's Formula. You misspelled Leonhard Euler's last name, which is surprising considering this is a math channel and he's one of the greatest mathematicians ever.

[@ferdinandrius6063]

Faux car a^z=e^(z*ln a) par définition. Donc 1^z=e^(z*ln 1)=e^0=1.Vérifiez avec hp prime , mathworks , geogebra, xcas, maple, wolfram alpha et j'en passe ! piège à clic bien sûr !!

[@l.w.paradis2108]

This has no solutions, but doesn't it have a limit? Lim (2^(1/x)) as x approaches infinity = 1. Is this correct? Wouldn't that be true of the xth root of any number, as x approaches infinity?

[@v2talk]

Not for 1 as 1 raised to any real power remains 1

[@yiutungwong315]

1^All Real Number = 1

[@l.w.paradis2108]

@@v2talk But no one is questioning that.

[@marcusgloder8755]

1^(x) = 2 x = log₁(2) x = infinity

[@akulkis]

Not even infinity solves it. 1^n = 1 for all n

[@vasiliybr9510]

X это К ≠ 0???? Заебись ответ. Стоило полчаса выдумывать ничто из ничего. То же самое я могу сказать, что бытие двери это ничто бытия окна. Вы мне просто скажите: какому числу равно x!?

[@Hayet-jb2sd]

Est ce que 0 peut etre un denomerateur?

[@KrekFret]

Où avez-vous vu le 0 dans le dénominateur ?

[@kuuso1675]

Is complex-log well-defined?

[@danielbriggs991]

Excellent question. There are several possible definitions for complex log, and in each case it is well defined, but since different texts use different definitions, it's dangerous to assume one definition in another text, or sometimes even in another situation in the same text. The most common in my experience is to give it a branch cut across negative real ray, and group that ray in with quadrant 2. That is to say, ln(re^(iθ))=lnr+iθ *for -π

[@kuuso1675]

@@danielbriggs991 Thank you so much for your clear explanation. Then in this problem case, that no real solution exists leads that we may choose the real >＝0 axis as the complex-log branch and formuli hold true in the left well-defined domain. Right?

[@danielbriggs991]

Yup! It's a very common tactic if you want something to hold about log in an open neighborhood of a negative real to use a branch cut along the positive real axis.

[@stehbar]

outrageous

[@woodbell3001]

I had no idea it was Euler's formula.

[@jamesnapier3802]

The guy's name was Leonard Euler, not "Eurler".

[@robertriachi6396]

Hy. Sorry to tell you that when verifying 1^x , the result is 1 and not 2 !!

[@jaju2178]

His name is EULER.

[@THyperon]

Euler's, not Eurler's

[@mentalitydesignvideo]

I think it was pretty obvious right at the start that there's no solution.

[@haroonmed9959]

If k is in Z then -k is also in Z

[@samueladler9080]

If that is x, then test it by raising the solution to the value whether real or complex, and see if it would return 2

[@jonathanr520]

There's no solution because no matter how many times you multiply 1, you cannot not achieve a value of 1.

[@peterbudai7415]

this also was my first impression

[@IAmTheNiceGuy]

there is no real solution. you are using the intuitive way of thinking, which i admire by the effort. but when we say something like 2^(1/2), this is not 2 multiplying itself half times. Math is abstract, 1^complex number can be something different than 1.

[@steve25125]

*1ˣ = 1, ∀x* *1ˣ = 2, ∄x*

[@gisa2961]

Magic lesson

[@SmailMoudoub-yv7yx]

Error because you have supposed cos◇=1 when k=0. There is contradiction.

[@bluestrue]

Dear friend, 1 raised to the power of any number is 1. Thus, LHS becomes 1 and RHS becomes 2. In other words, 1 is equal to 2, which is fallacious. If we take logarithms to the base 10, x log 1 is equal to log 2. Simplifying, x is log 2 divided by log 1 which is undefined.

[@SMARTpeople1135]

오 재밌당❤

[@maurizioomissoni8446]

It's Euler not Eurler

[@timcoale160]

Waste of paper and ink

[@james2578]

Just like ur a waste of ur mothers egg.

[@PROTAEQUESO98]

r/rareinsults

[@alphastar5626]

The whole channel is like that fr I don't mean to be rude but the dude should really consider doing something else in his life

[@nickkunst952]

Why? Because you cannot understand it?🐒

[@alphastar5626]

@@nickkunst952 yeah right whatever. It's a maths for teenager entertainment channel so i guess he got his target right..

[@kumalala136]

Impossible

[@harrymattah418]

Eurler... OK bye.

[@Bogdmih]

Формула Ейлера працює лише на відрізку від 0 до 2*пі. Тому не потрібно маніпулювати, оскільки 1^x=1 завжди і всюди!!!!!!

[@Ostup_Burtik]

1^∞=кожне число

[@Bogdmih]

@@Ostup_Burtik Що ти хотів цим сказати???

[@Ostup_Burtik]

@@Bogdmih нескінченний корінь із x =1, тобто 1^∞ будь яке число. не потрібно 500 знаків питання.

[@Bogdmih]

@@Ostup_Burtik І що ж далі???

[@Ostup_Burtik]

@@Bogdmih знову 5000 знаків питання. Нічого далі. Просто 1^∞=будь яке число

[@ゆーりん-k6u]

これ間違ってるよ この前の解き方も間違ってたし

[@nostrawneeded]

I was searching for a bfb video

[@judecloud]

눈이 번쩍 뜨게 하는 풀이네요. 감사합니다.

[@manishasahu2924]

x=x+1

[@tontonbeber4555]

There is no solution real complex quaternion or whatever set you wish.

[@IAmTheNiceGuy]

you are wrong, for U=C, there is solution.

[@Rossbach2]

1^x=2 Log 1^x=Log2 x∙Log1=Log 2 x=log2/log1 x=0.301/0 x is undefined.

[@jamesharmon4994]

It's pronounced "Oiler's formula".

[@ivocosmi5303]

Esta mal, sustituyendo ese resultado en el problema de ecuación no se cumple la igualdad; un ejemplo en las ecuaciones de segundo grado que tienen raíces imaginarias, al sustituirlos en la ecuación se cumplen la igualdad, pero en este problema no cumplen la igualdad; Por ésa razón esta mal.

[@SIB1963]

FTR, "Euler's" is pronounced "Oiler's".

[@mijokalebic4371]

Eurler😂

[@pasixty6510]

I comment this one only for the sake of sympathy. I am as smart as I was before the video. Only a few minutes older. Sorry, I didn’t get it…

[@ivanyeung4577]

not completed answer, tell u later

[@岸辺緑]

0以外のあらゆる複素数を0乗すると1になる。故に、この動画の論法を認めてしまうと、1の冪乗は任意の値にできる。 要するに、この動画の内容は出鱈目。

[@yiutungwong315]

0^0 = 1 0! = 1 1^All Real Number = 1

[@F007-n6y]

1ˣ=2 х=log₁2 x∈∅ 1∉(0;1)∪ (1;+∞)

[@Bullit7676]

If that real then x=log 2 1

[@IAmTheNiceGuy]

it isnt real, this is in the complex numbers

[@haothe1099]

1log = 2

[@Hayet-jb2sd]

La simple kima kolt

[@GTSpeac]

Is this question made up? - for any 1^x there can be no value other than 1 irrespective of positive, negative, real or even 0.

[@IAmTheNiceGuy]

1^x can be any number in the complex plane, except for zero, we dont really know any form of solutioning for 0, but for the others, we can!

[@GTSpeac]

@@IAmTheNiceGuy irrelevant - the answer would still be one :-) - otherwise bring your proof.

[@IAmTheNiceGuy]

@@GTSpeac ??? the proof is literally on the video! you are trying to appeal to common sense to deny the proof, but common sense only applies to elementary math, when we come to the complex world, it is totally possible to deny things like 1^x=1 because it isnt an essential axiom of the number group.

[@GTSpeac]

so what is x in the context of 1^x=2?@@IAmTheNiceGuy

[@venkatesank3841]

This question is meaningless. Log 1 is 0. Log 2 is 0.3010. And not 0.693. Any power to 1 is 1

[@stanisawk1385]

No, tu już się popisałeś! Piszesz takim samym znakiem czyli kreską zarówno cyfrę jeden jak też literę L. To bełkot. To jak napiszesz wartość bezwzględną z liczby jeden? Ja ci to napiszę: |||. Taki z tego wyszedł bełkot i bezsens, ponieważ NIE CHCE CI SIĘ pisać poprawnie liczby jeden, z ogonkiem u góry. Tak jest poprawnie: |1| Gratulacje

[@Andrew-WK]

Как вы любите усложнять... Есть правило, на которое все опираются: 1 в любой степени это 1 Поэтому нет решений. Остальное - какая-то хрень. Всё, спасибо, расходимся

[@freudefreud]

ты хоть среднюю школу смог закончить?

[@Evgeny_Miroshnichenko]

@@freudefreud А ты?

[@electron252]

Не так все просто. Можно выразить решение комплексным числом, что и показывает автор. Кстати, в нормальных школах еще знакомят с комплексными числами, а по повторительному учебнику Сканави "Элементарная математика" можно понять, что в 1970е это также было общераспространенной практикой.

[@XyxpbI-MyxpbI]

1 в вещественной степени имеет единственное вещественное значение равное 1. А вот комплекснозначных различных, в том числе и комплексных, степеней единицы полно. И среди них неожиданно появляются даже такие, которые равны в том числе и 2, хотя, казалось бы, 2 не просто вещественное, но даже целое.

[@Vadim1488]

Это аксиома или вывод из аксиомы?

[@AndrewWutke]

There is no solution whatsoever 1^x=1 hence 2=1

[@Hayet-jb2sd]

Je n'ais pas compris?

[@chinmoychakraborty6116]

Unreal equation

[@Robobocs]

0,5

[@GerryFolf]

Можно было взять определение степени? Ну там, степень это краткая запись умножения числа самого на себя. Откуда следует, что 1*1*...любое число раз...*1 = 1. А не лепить абы чего через какие-то логарифмы, дабы показаться умненькими. Но "математики" никак не угомонятся поэтому лепят дальше абы чего.

[@СергейСпиридонов-ш5н]

Из всего решения понял, что там какой-то "икоту" есть

[@sammail180]

ответ 0 - считаем устно!

[@DmitriyObuhov]

Решите такую задачу Х/0=2

[@ewerest9914]

2Kpi means 0 degree. You going to divide by zero

[@sytrostormlord3275]

Only if k=0... 2pi = 360 degrees...or ~6.18 in Rad... so you're dividing by integer multiplications of those

[@ewerest9914]

@@sytrostormlord3275 Why then dividing by zero is mistake? You could replace any zero to 2Pi. Euler's formula use the degrees, excetly the degrees. 2Pi degrees is not the same as 2Pi of length. 2Pi degrees =0 degrees. Of course you could rotate smth by 2Pi degrees, but there will be two state - start and stop and moving (trip = 2Pi*r) between. One math expression describes one state or an instant, not two. Euler's formula and complex numbers in general have been using in and inventing for all ingenireeng calculation where being degrees. The reason of inventing complex numders is the degrees...

[@Jan_D-vm5rk]

No, that is not true. Euler's formula uses numbers and not angles, and mathematically the functions sin(x) and cos(x) are defined for numbers, independent of angles. So mathematically sin (π/2)=1 and mathematically sin(90) ≠ 1

[@Highestknowledge]

Suddenly write 2 senseless

[@andrewauh2605]

Waste of time I'd say. Any power of 1 is always 1. The complex solution was not a solution.

[@LouisVanWinklaar]

*x = -2*

[@SD-it5ko]

5:40 You can't do that with complex numbers

[@Андрей-я7и1с]

Как же достал этот "умник " решения изначально нет . Написал охинею и другим мозг засоряет.

[@galinaapostolyuk7977]

Якщо вас він достав, то чому дивитеся відео? Можливо математики придумали комплексні числа, щоб ми нічого не розуміли.

[@chandreshsaroj30]

Hiiii

[@learncommunolizer]

Hello 👋

[@aqlimursadin5948]

It's a scam

[@katakgemok]

why!!!!!!

[@garungarun8233]

e^(i((л+2кл))=-1 1=-e^(i(л+2кл)) (е^(i(л+2кл)))^х=2 x(i(л+2кл))=ln2 -i ln2 X= ---‐--------- л+2кл and you - i ln2 X= ------------ 2кл ???.

[@garungarun8233]

x So 1 = -2 then -i ln2 X= ---‐--------- л+2кл

[@Hayet-jb2sd]

By by

[@walterpcjr]

Absolutely wrong!

[@abtinnavid6903]

This is not correct!

[@petechen794]

No real solution

[@dimitrovniko608]

1^0 = 2^0 So, the answer is x=0

[@andrewtch-hk]

1^x = 2, not 1^x = 2^x

[@mrmimi807]

The equation is wrong so you can't apply any rule

[@daddykhalil909]

A very trivial and unnecessarily complicated problem You jumped from real numerical world to some imaginary space applying theories and ideas which are not known to most of people Unfortunately your work is a pure loss of time and efforts time

[@usafadonnkzyalog6]

Bruh