Can e^e^x=1? Sol here kzfaq.info/get/bejne/mdGTbKlhu9C-Y58.html

“360, but we are adults so we use 2pi” I felt that

Imaginary numbers are the math equivalent of going into the shadow dimension to get through obstacles

Notice that 1^x = 2 and 1 = 2 ^(1/x) are, actually, two diffferent equations with different domain of x. You solved the second equation and not the first one. Edit: that is EXACTLY why wolfram can solve the second one.

Actually Wolfram-Alpha is correct. Too understand why we will need some function-theory/complex analysis (for example: Complex Analysis, Elias M. Stein S. 97-100). At first we will need a definition of z^w with w,z in C. For any z in C\(-∞,0] we can define a function z^: C --> C by z^w:=exp(log(z)•w) where log is the principal branch of the logarithm (that means that log(1)=0). Of course you can choose another branch but in this case the definition does not match with the exponetialfunction with a real basis. Using this definition we get: 1^x =exp(log (1)•x)=exp(0•x)=1 which states that the equation 1^x = 2 got no solution. Now we take a look at the Question: "Can we finde a x in C such that 2^(1/x)=1?" Using the definition we get 2^(1/x)=exp(log (2)•(1/x)) which is equal to 1 whenever log (2)/x=2πi•k, for any k in Z. This gives the solutions you are getting too. After clearing this we should talk about the "contradiction" at 06:23. What you are writing there is correct but its not a contradiction to the above: 1^x=2 => 1=2^(1/x) means "every solution of the first expression is also a solution of the second Expression" (which is correct cause the left expression got no solutions). The other direction 1^x=2 x²=1 is correct but x²=1 => x=√1 is wrong (the solutions -1 "gets lost"). And this gets even "worst" when complex numbers are involved...

When blue pen gets in involved, you know it's serious...

"Can 1^x = 2?" - No. *video ends*

I like how this whole time he was holding a poke ball and half of us were too busy having our brains crushed to realise

“We need two things. The first thing is the distance. The next thing is to erase the equal sign better. The third thing is the angle.”

That felt like a mathematical crime.

Mathematicians whenever they wanna look complicated : Let's talk about complex numbers Physicists whenever they wanna look complicated: let's talk about Quantum physics. Chemists whenever they wanna look complicated : let's talk about chemistry!

My man starting to look like an ancient philosopher who lives on a mountain, I dig it

"I don't like to be on the bottom, I like to be on the top."

This is my equation of the year in 2021. To see others, please check out here 👉bit.ly/equationoftheyear

After finally having learnt complex numbers, it feels so good being able to understand these types of videos! Keep up the great work.

I would guess 2 possibilities: 1. No. 1^x cannot equal 2. 2. If 1^x can equal 2, then 1^x can equal anything, because there is nothing special about 2.

You are evolving into one of these chinese big beard philosophers lol

Man, you are literally bug hunting at this point, they should be paying you. Very nice video

This is very similar to the equation sqrt(x)=-1. If you put that into Wolfram, it will tell you that it has no solutions. You can try to argue that well, actually, one of the square roots of 1 is -1, but the thing is that's not what sqrt(_) actually is. The same is true under complex exponentiation: the principal branch is used by definition and as such, 1^x=1 no matter which x you plug in. As others have pointed out, this does not contradict that 2^(1/x)=1 does have solutions in C (even when we are taking the principal brach). So no, Wolfram's right here.

He went from: guys I have pen and I do math to: 筆子曰：「無實數解既找虛數解」。

Can 0^x=2?

"ok so we'll just take a logarithm and set the base to 1" this triggers me

Okay if you solve for x in 1=2^(1/x) using wolfram alpha, you indeed get x=-i*ln(2)/(2pi*n), but as I have learned in math competitions, you should -always- usually plug it in to the original equation. And plugging x=-i*ln(2)/(2pi) as 1^x in wolfram alpha, you get 1. So x=-i*ln(2)/(2pi*n) are all extraneous solutions, which is why they are not solutions to the original equation 2^x=1.

Answer from google: Logarithm is not defined for base 1.

The math police have issued a warrant for your arrest.

You know what. This came in my recommended, and man let me tell you, my algebra 2 teacher must be doing a great job because I don't know how I willingly clicked on this and was genuinely interested lol

I love this! I'm taking an EE class right now which revolves around complex numbers, and your videos are super helpful/inspiring.

I think wolframalpha is right here. 1^x=2 has no solution but 1=2^(1/x) does. Grinding this down to fundamentals you see that (a^b)^c is not equal to a^(bc) for complex numbers, exactly because the change of branch of log you expertly portrayed in the video. Another instance that messes with this is my all time favourite: a = e^(log a) = e^((2\pi i / 2\pi i ) log a ) = (e ^ (2\pi i) ) ^ (log a / 2\pi i) = 1 ^ (whatever) = 1.

“Yes, but not all the time” YOU’VE BROKEN MATH

2:59 I was gonna say 360 degrees like the child that I am. I can't wait to be an adult and say 2 pi!

There are many things wrong with this video, but they serve as good indicators of why we need the concept of branches in complex analysis. Let us first see that some things are certainly not right: Suppose the reasoning by BPRP works, and that one can indeed write ln(1) = ln(e^(2pi*i*n)) = 2*pi*i*n, for any integer n. Then we run into the problem that ln(1) = ln(1^2) = 2ln(1). This immediately implies that ln(1) = 0, because that is the only solution to x = 2x. So either BPRP is wrong, or ln(1^2) = 2ln(1) is wrong. However, this same property of logarithms is used by BPRP himself when he writes ln(e^(2pi*n*i)) = (2pi*n*i)*ln(e), thus in any case BPRP's argument is not self-consistent. This property of logarithms should be familiar and we would certainly want this to be true. Let us now get to the heart of the problem. BPRP did not consider that in the case of inverting the complex exponential, you may not use all properties that we are used to when dealing with logarithms of real numbers. To 'invert' the complex exponential, you need to choose a specific branch, precisely to deal with problems like the one that we see in the video, namely that 1= e^2pi*i = e^4pi* i = e^6pi*i = e^(2pi*n*i) for any integer n. What does choosing a branch mean? Well from these equalities we see that there is no single inverse value to the complex expontential for the value 1: We need to choose one of the values 2*pi*n*i to get a step closer to defining something like an 'inverse' function to the complex exponential. Making such choices in order to always know which value to pick is (crudely speaking) what mathematicians call choosing a branch. The natural logarithm for complex numbers is an example where such a choice of branch has been made: The natural logarithm is defined for complex numbers by choosing the principal value branch, which restricts to the interval (-π, π]. This means that even though 1 = e^2pi*n*i for any integer n, when we use ln(e^(iθ)), we choose the value inside (-π, π] (even if θ is outside this interval!). In the case of ln(1) = ln(e^(2pi*n*i)) for any integer n, the natural logarithm then simply gives 0. This last point stresses that ln(e^(2pi*n*i)) = 0, thus the argument in the video does not work. Then one might be tempted to defend BPRP's argument by saying that he implicitly chose a different branch for the natural logarithm, precisely by asserting that ln(1) = 2pi*n*i for some integer n other than 0. However, even then one encounters the problem we discussed above: ln(1) = ln(1^2) = 2ln(1). This does not mean that one cannot take different branches for logarithms. Instead it means that when we do take a different branch, we cannot expect precisely the same rules to hold for our logarithm. In particular the rule for logarithms that log_a(b^c) = c*log_a(b) does not hold anymore if we choose a branch corresponding to ln(1) = 2pi*n*i for n ≠ 0. Of course this is a cherished property of logarithms, and motivates even more why mathematicians prefer to choose the principal value branch: It is the only branch in which this property holds. Thus in conclusion, BPRP's algebraic gymnastics to solve the equation 1^x = 2 is not correct, and upon further inspection Wolfram Alpha turns out to be exactly right: There is no solution to this equation. However there are some solutions to the equation 1 = 2^(1/x), which is NOT an equivalent equation. But my comment is long enough as is. If anyone is interested, I can elaborate more on this later.

Question: Do you need the negative sign in the numerator? If n runs through all positive and negative integers, it's the same "solution set" with or without the negative sign ... ... isn't it ?? PS - Your gentle emotional delivery here (moments of disappointment, exasperation, near-defeat) is a refreshing new contribution to the art of math videos!

2^x=-1 vs. (-1)^x=2 but in ONE minute kzfaq.info/get/bejne/pqiehpp8ktumXZs.html

As a random 13 year old, my mind imploded and exploded at the same time

Moving math into higher or adjacent dimensions is always a very neat thing to see. Things you think that are impossible become commonplace

Dude your flicking motion to flip between the markers is so smooth.

bprp: 1^x=2, x=? Wolframalpha: *By assuming x€R,...*

You have just extracted juice from a stone. It´s beautiful.

What a great video, the ending really makes it all the more satisfying.

Great Maths! I'm currently in my first year of A Level Maths and I felt like I could follow it

Wait a minute, Log base 1 is undefined Anyways, The pokemon in his hand is more important

Thank you. This is one of the cases my Math Prof. warned me about. Does the imaginary unit really enable such relations or are the exponential / logarithmic laws more limited in the complex world than one might first time think?

Cancelling log and exponent so casually gives me anxiety about messing up with branches....

I came here because I was curious of how you would have solved the equation, to discover that behind it there was a story of a change of perspective. Great video, I had fun, even though I can't do all that math :)

This answers a question I've had for a long time: does some math treat e^2 and e^(2+tau*i) differently? I learned something today. Thank you so much :D

Teacher : 1 to the power anything is 1 BPRP: Hold my M A R K E R

So thorough! Thanks a lot!

The most EPIC beard in all of the KZfaq Mathematics community! Also, LOL at that look-of-disappointment at 2:31! 😆

The first thing I noticed is he don't have glasses.

😂😂😂 omg. I love the fact that you're searching for interesting equation to solve. That's amazing 👏🏻 keep going . This is your folower from Morocco ❤️

*blackandredpen: writes log1(2) to the board* Me: *wait, that's illegal!*

I think this is probably my favorite problem you've done. It just shows how completely messed up things can get when you get to the complex plane, to the point where 1 raised to a power does not equal 1. Complex analysis is just bizarre. But in the end, this problem is so dang easy. You can get to things like -1^x = 10 x = i ln 10/pi How crazy is that?

All depends of the branch of logarithm chosen because k^x=exp(x*"log"(k)) where you need specify the 2*pi magnitude interval of the imaginary part of function "log" how is defined; if it contains 0 there are no solutions. For example, the case of principal Log doesn't work because Log(1)=0; thus, for powers of principal branch 1^x is always 1

wait, that's illegal!

Fascinating question, hooked right away

very good calm explanatory teacher

One of the first videos I was ahead of. Guess studying is paying off.

Depend on which branch of 1^x you take

this man can bend reality

Sir, I’m going to major in mathematics at uni, and your videos both make me realize how terrifying that is but also somehow reassure me.

All I could see was the beard.

another way to solve this is 1=i^4n, n being an integer different than 0, you then have ln(i^4n)=4nln(i)=4nln(e^pi/2*i)=4n*pi/2*i, so ln(2)/ln(1)=ln(2)/4n*pi/2*i=ln(2)/2n*pi*i=-i*ln(2)/2n*pi, same result with a slightly different method

Love your videos

i love how he explain in a way that i seem to understand, but then i realized i have no clue on what he's doing

Is base 1 for logs exist?

I came here to see how many people undestand this, because boi i really weak in math. It took me hours to understand this, and open bunch of books about ln, log, and how they works. By the way i'm in 8th grade, and your vids helped me to understand lots more of how to calculate. So i thanked you for that 💙

You are literally a hero!!!!

I hope your videos never disapear.

When Wolfram alpha breaks you know your fkd.

BPRP without glasses is triggering my fight or flight response

You know the video is going to be good as soon as you realize that the teacher is about to start a Pokemon battle

Do you have any examples of equations that can be solved using quaternions only ?

Math itself must've felt violated after the problem was solved

4:15.....why would he assume that WE know "HE" likes to be on the top rather at the bottom?

You blew my mind🤯

Nice video ! And nice shirt !

A complex number raised to a complex power has multiple values. If we treat 1^x as such a multi-value expression, and the question is to find x so that 2 is one of values of 1^x, then x=-i*ln2/(2*pi) as stated in the video is a solution. Actually there are multiple solutions: x=-i*ln2/(2*pi*k) with k being any non-0 integer.

The title should be "10 ways to (not) write zero"

thank you for this video!! since the 7 of january is my birthday so this is a good gift!!!

Yes, if x = (-ln(2)*i)/(2nπ), where n is any non-zero integer. But only for non-principal values of 1. The principle value of 1 is e^(0*i), or e^0. And raising e^0 to any power will yield e^0, which is 1. So entering 1^x into a calculator will yield a result of 1, even if you use this x value. But if you use this value of x when entering 2^(1/x) into a (sufficiently sophisticated) calculator, you will, in fact, get 1. EDIT: Cool, I didn't miss any details. I actually solved this slightly differently from how you did. I actually jumped straight into polar form, restating the equation as (e^((0+2nπ)*i))^x = 2 right off the bat.

Starting the day with these kinda sums as a jee aspirant feels refreshing.

I have no idea why I'm watching this on a Saturday evening, but here I am

Great! Always someting surprising.

You're a great teacher :)

log functions are not defined for base 1 . in fact its in the definition that base of log is positive and can't be equal to 1 . so how can you take log with base 1 on both sides 🤔

bro chill my girl’s on this app

You can change ln(2) also to ln(2e^(i(0+2pi*m))) This way you can get more answers X=(ln(2)+2pi*m*i)/(2pi*n*i)

You can put 1 = i^4 in ln(1) or in initial equation to solve in a different way

This proves it, complex numbers were invented by mathematicians who were on some extremely good weed.

Hello there is a mistake from the beginning when u wrote X=log(2) because log(1)=0 so basically X is multiplied by zero

Brilliant and beautiful. In this example, the creative thinking process is demonstrated (or even exposed). How to achieve new results.

In general, if 1^x = z, with x,z ∈ *C* and z≠0, then x = k + [(ln z)/(2πm)] i for some k,m ∈ *N* with m≠0. In this case, the values of 1^x are the principal values of z^(-n/m) for all integers n. So the principal value of 1^x is always z^0 = 1, but the desired value is always in there. Specifically, 1^x = z on the branch n = -m.

How about "x tetration i = 2" :0

The pokemon in the pokeball in his hands probably learnt more maths than me

This is an excellent video, thank you very much, greetings from México

I'm a bit confused. No calculator that can handle Complex Numbers gives the Result of 2 for the formular with n=1. Does this have something to do with the Complex-Log having multiple solutions? Or am I misunderstanding, what he is trying to say?

4:37 My brain: starts melting

Why tf is he holding a pokeball is he trying to get sued by nintendo

There should be more solutions. You can also split 2=2*exp(2*pi*m*i) and use the additive rule of the logarithm. The general solution is x=-iln(2)/2/pi/n+m/n,n is not 0,m,n are integers.

Your vedios are good..... I am inspired bye you in mathematics..