Is e^x=ln(x) solvable?
ฝัง
- เผยแพร่เมื่อ 25 ก.ย. 2020
- We will solve an interesting algebraic equation involving both exponential and logarithm, namely e^x=ln(x). Although the graphs of y=e^x and y=ln(x) do not intercept, we can actually find some complex solutions to this equation. We do need to use the Lambert W function tho. So see here for a detailed lecture. Lambert W function Lecture: • Lambert W Function (do...
We will make b^x and log_b(x) tangent to each other here: • the famous equation b^...
Check out Mu Prime Math's video on when is f(x)=f^-1(x)=x true: • When does f(x)=f⁻¹(x) ...
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We will make b^x and log_b(x) tangent to each other here: th-cam.com/video/uMfOsKWryS4/w-d-xo.html
"We are not doing real mathematics."
-blackpenredpen
It looks like pretty real mathematics to me. I have always thought real and imagery numbers both exist. They are both items in the study of mathematics. Therefore these labels are not really suitable.
@@thomaskember4628 yeah people start using the term complex numbers instead, but imaginary do be sound cool so
glacifiess A complex number is not the same as an imaginary number, it has a real component. When I was learning mathematics at school, I thought imaginary numbers must be the lest interesting part of mathematics because as soon as they are introduced, we go on to complex numbers.
The ee... = e(ee...) begining assumption is Wrong Infinite series ! Thus the final a+bi derived from Wrong Assumption violate the complex definition ! Thus Real=Complex , x=z give
people people the wrong idea/logic mixed up ! Similarly you can prove Girl = Man exactly the same ! ! !
@@klong4128 Are you okay?
The solutions for this equation were way too complex that i couldn't even imagine
Well, I couldn't either : )
( :
One of the solutions to this is infinity because e^inf=inf and ln(inf)=inf
@@dannyyeung8237 actually no because this is in an indeterminant form
@@dannyyeung8237 Wow this is so terribly wrong
“We are not doing real mathematics. We are doing complex mathematics.” I need that on a shirt
One of the solutions to this is infinity because e^inf=inf and ln(inf)=inf
@@dannyyeung8237no
@@dannyyeung8237thats alos what I thought with e^e^e^e^… which divergences to positive inf.
@@dannyyeung8237infinity is not a number. It's just a placeholder for "impossibly large". It's just something that numbers approach but never really reach.
All what uve done proves is that both functions diverge as x approaches infinity. But lots of functions diverge. We wouldn't say all functions that diverge "equal" eachother when x is infinity. What is meant by divergence is that it just keeps growing endlessly without limit the higher u increase x.
U can also apply the same logic to the equation 2x = x+ 1.
@@vwlz8637Completely depends on the number system you define your function on. There’s no value for infinity in the reals but there is one in the extended reals (i.e. ℝ ∪ {∞})
Note: The following equations have the same solutions!
1. e^x=ln(x) *this video*
2. e^x=x
3. x=ln(x)
4. e^e^e^...=x *this video*
4. x=ln(ln(ln(....)
This is a super nice property when you have f(x)=f^-1(x). See Mu Prime Math's video for more details: th-cam.com/video/53lBKCBrENY/w-d-xo.html
I tried an alternate version of this problem by proposing the following set of equations:
1) exp(x)=ln(x)
2) exp(-x)=ln(-x)
Combining both equations, I ended solving sinh(x)=2πi (I am not considering all the logarithm branches in the complex world, I just picked ln(-1)=-iπ) and my final result was:
x= ln{[π+-√(π^2-4)]/2} + iπ/2
Is that correct?
P.S.: Greetings from a big fan in Spain.
#YAY
Drop the bomb
I thought I got clickbaited when I heard you say we're not doing real mathematics. Lol
Would this work tho? Because since it was e^e^x it would always be an even number of e's. So the replacement to convert it into e^x would include the solution, but it could also include more solutions.
I guess this is salvaged by the fact that only one solution was found
black pe^e^e^e^e^e^...n red pe^e^e^e^e^e^...n
«Infinitely many e's...wow!»
@@jagatiello6900 but if there is infinite amount of e's then he would never get to the letter n HMMMMM
Black peen red peen
Simple math:
BLACK PEN
RED PEN
Complex math:
BLACK PEN
RED PEN
BLUE PEN
.
.
.
even more complex math:
+Purple Pen
@@Killer_Queen_310 quaternion
3:21 wow in australia.
India 12:55
˙ɐılɐɹʇsnɐ uı ʍoʍ 12:3
I sent the same in an email to a friend at the same moment.
Your biggest fan from Russia! Love your videos so much, you're my best math teacher (and English too:)) since 2018, thank you!
Thank you! I am very happy to hear this! : )
здрасьте
о я тоже смотрю его с 2018
помню было весело когда приходилось на уроках математики переводить его речь в голове на русский
@@hiler844 доброго вечерочка)
@@redblasphemy9204 я к тому времени уже в универе учился, так что видео были кстати)
Let me guess, it involves the lambert w function
I was going to say there would be fish, but unfortunately, they weren't used today.
@@particleonazock2246 "Have you got any fish?"
"Go fish!"
Fred
@@ffggddss "Then he waddled away~ (waddle waddle)"
@@particleonazock2246 Just put the lambert W function. And you got your fish BACK.
i commented my own solution- take the equation at 1:58 and multiply by e^x. take W on both sides to get x=e^x. then solve
It's only through this channel that I learned of the Lambert W function and have become fascinated, wondering why it was never mentioned in college-level calculus.
Why it should be mentioned?
I'd never heard of it before either. Upon looking into it more, the reason is probably that
1. It deals with complex numbers which generally isn't covered until later in college
2. The applications are really niche and not something a student in college algebra would use until way later if at all
It's quite artificial, imo. Does come up naturally that often, but rather usually in these sorts of intentionally awkward constructions.
I think it was mentioned/used at some point in my linear algebra class, though. Would need go trawl through the notes
@@Cyrusislikeawsome it comes up plenty of times in physics. First thing that comes to mind is that it is part of Wien's constant (when solving for the maximally emitted wavelength of a black body).
@@jorenheit I'm p sure that's the one of two occasions I had in mind aha. Genuinely, any more?
Reals: Nope..no solution to this thing..
Complex numbers: Hold my “i”s
Once you get
x = e^(e^x)
Can you multiply both sides by e^x?
That would give you
xe^x = (e^x)e^(e^x)
Then you could use the Lambert W function on both sides to get
x = e^x
That last step doesn't work. The Lambert W function isn't single valued so you only get that any solution of x=e^x is a solution of x=e^(e^x) (which is something you could have obtain in an easilier way) but not that all solutions of x=e^(e^x) are solution of x=e^x
Check pinned comment equation 2
@@mathieuaurousseau100 The question is to "find a solution"...
@@emeraldng2910 But the title was "solve"
More importantly, there can be cases where f(f(x))=x has a solution but f(x)=x don't
Test: solve e^x=ln(x).
me: "eeeeeeeeeeee...."
I am glad that you showed the Lambert W function. I have learned QFT and statistical mechanics for sometimes, but I didn't know about the W function. That will be helpful.
I took interest in maths after watching your videos.
Stay safe
love from Nepal
me too from nepal
i stared at the thumbnail for five minutes thinking before clicking. this strategy works
3:34
We are not doing Real Mathematics PANIk
We are doing Complex Mathematics kALM
We are doing Complex Mathematics PANIk
But it's bprp kALM
Awesome video as always!! I started watching your channel years ago and now you are really helping with my Further Math lessons (I’m from England). If I have a problem to suggest to you, where can I submit it?? Thanks again for a great video!
You can always email him, he responds to those sometimes. Though since you posted this comment last year, I imagine you've finished your Further Maths A Level by now anyway. I've also been doing it the past two years, and I'm relived to have finally done the exams, hell as they were xD. How did you find them yourself?
What I was looking for, a video for the complex natural log. Thank you so much B&Rp!
I loved the 🐧,the song and not to mention "we are not doing real mathematics"✌🏻😌
Wait... if e^e^e^e^... doesn't converge, is it valid to say
x = e^e^e^e^... x = e^x
?
Not in the reals : )
When it doesn't converge you just don't consider it a solution just as you would do with an imaginary number while working in the reals
Is there such a thing as convergence in the set of complex numbers?
@@ekeebobs7520 of course there is
@@ekeebobs7520 dependent on what your question is:
- Is there a concept as convergence in the complex numbers? Yes! if you know epsilon delta/N definitions you can now interpret the absolute value as the complex modulus and this new definition still makes sense. for example, it is a theorem then you can then say lim (xₙ+iyₙ)=lim xₙ +i lim yₙ. that way it is easy to continue intuition and also taking limit of sum is sum of limit etc is still true this way.
- Does this converge in the complex numbers? absolutely not still. simply because the sequence e,e^e, e^e^e,... has a very fast growth
first time heard "lambert W function".
This is one of my favorite channels!
This was helping with extending tetration to complex numbers. I remember Dmitry Kouznetsev mentioning this as fixed point of logarithm. I guess it can be found by trial and error but obviously there are other methodes.
I miss the intro from 2017: " Black Pen Red Pen yaaaay!" Like if you agree
@Tropical_Papi It was cringe.
@@adityakamat9856 no
If you treat x as a+bi, and use the (equivalent) e^x = x equation, you can show that there are an infinite number of solutions on the complex plane. The solutions are all in Quadrant I, asymptotically approaching the curve b = e^a, where b takes on the values of pi/2 + 2*pi*n for large n.
Watching from South Africa. You made me love math man! Keep feeding me
No comments? :(
why is it unlisted?
2 months ago ??????
How
What dude??? 2 months ago???
How man
Who are you ? **meme insert**
2 months ago? hax
My first thought was that if e^x = x, then ln(x) must equal x, which necessarily means that e^x = ln(x). So we can start by writing the question as e^x = x, and then use the W function.
Before: Black pen red pen.
After: Here comes blue pen.
Ah... Finally, come back with another video. ❤️❤️
Black pen red pen you are the best at what you do
when you reached x=e^e^x you could simply multiply e^x on both sides and then take a lambert w function and you would end up with x=e^x.anyway.it was a flossy video:)
Ah. Good point.
Yes you are right!
I wanted to show the e^e^... part and that’s why I continued 😃
or you could work backwards and see that you could get to x = e^e^x if x = e^x by substituting for the x in the exponential
One of the solutions to this is infinity because e^inf=inf and ln(inf)=inf
Wait the link works
Could have done with a demonstration of why an infinite tail of exponents is allowed, when it appeared that it would have to terminate with an x power after a finite and even number of e terms.
that last sound really makes my day!
Can you maybe so a video on how to evaluate the infinite power tower? I guess it's probably with some sort of a sequence and a difference equation, but I would like you to show us how to do that.
What do you suggest to prove it please
haha. like the way you make it simple to complex to simple. and infinite problems always are insane :)) nice vid.
I was never thinking there are fixed points for exp() (except for transfinite numbers). Nice to know.
these recursive thing is mind blowing... nice I get to learn new stuffs
AMAZING!! Your skills never stop amazing me!!
So if you plotted e^x and ln(x) in a 3D graph where z is the complex axis, would it look like two sheets that just barely touch each other at 2 points (at the two complex solutions shown)? Also, is there a cleaner representation of the solutions, or are both the real and the complex components transcendental?
If you’re allowing complex inputs you’ll get complex outputs. 2D mapping to 2D. You need a 4D graph.
Thanks for this video, congratulation since Mexico!!
It's very interesting! I don't understand English very well(?), but, the process you write on the whiteboard is very easy to understand!
"pause the video, and think, about, this."
"... and today we have this guy. ok so-"
😂
Always enjoying Watching your videos since when I was Grade 6, Nice Video as always! Love from the Philippines!
And yeah, I am now in Highschool :)
i love the sound effects
Love from India . I greatly admire your maths skills and teaching .
Jai Bharat!!!! 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
i would really love to see how to compute that lamber w function value at the end
you are really good at maths
Just learned about log and ln in class, this blew my mind
that hello at the beginning makes me happy:)
Love your videos
Very well done.
This morning, I was asking myself how could I solve ln(x) = e^(x), and I find your video the evening of the same day !
Holy crap ! I actually thought and tried to solve the equation and got the correct answer . Never thought could do it
when i was trying this on my own, at first i multiplied both sides by x to get xe^x=xln(x), and then took the lambert w function to get to x=ln(x)
The solution is re^ia where
r=a/sin(a)=e^[a/tan(a)]
Now this can be drawn on desmos and my god there's a billion solutions
Hey man love your t shirt
Note that bprp isn't wearing a mask, but his microphone IS - a full face..; no, a full BODY cover! Surely this puts him in compliance with Gov. Newsom's rules.
Not to mention the extreme social distancing he's practicing.
OK, first, replace x with z = x + iy
Take exponential of both sides:
z = x + iy = e^(e^z) = e^(e^(x+iy)) = e^(eˣ(cosy + i siny)) = e^(eˣcosy) e^(ieˣsiny) = e^(eˣcosy) [cos(eˣsiny) + i sin(eˣsiny)]
x = e^(eˣcosy) cos(eˣsiny)
y = e^(eˣcosy) sin(eˣsiny)
At first glance, I don't see where you can go from there. [I also tried starting with the polar form, and taking ln of both sides, which turned out even worse.]
Let's see how we can get anywhere with this. I smell the Lambert W function, somehow...
Fred
Love these complex solution videos!
Sir ur the best teacher of maths wish u were here in India to teach us ur teaching skills are amazing and u r the best................🙏
#JaiBharat!!!! 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
Fun fact for video game fans:
The mascot used for microphone
is from the Famicom game "Gimmick!" (released for NES as "Mr. Gimmick") made by Sunsoft.
This character in this game is one of types of enemies.
Awsome and clean
You know it gets complex when blackpenredpen includes a blue pen.
Nice work. I see that if we multiply both sides by x and solve we get
a. x exp(x) = x * ln(x)
b. Now apply the Lambert W function to both sides to obtain x = W(x * ln(x))
c. Now if we assume we can use the following identity, W(x*ln(x)) = ln(x), so the last result becomes x = ln(x)
d. We can solve x = ln(x) using the Lambert W function to -1 =- ln(x) * exp(-ln(x)) --> -ln(x) = W(-1)
e. x = exp(-W(-1)) = W(-1)/(-1) = -W(-1) or x = -W(-1)
I need a breakdown of the W function!
How it would be grafically with the complex answer?
Thanks for the videos! Keep going!
could you do a video on euler fonction, I have encountered it many times but still don't understand it.
You should try to solve a^x=log_a(x) where a>0. The critical value is a=e^(1/e). For a greater no solutions, for a less two solutions.
Best microphone!
Thanks 😆
Cool video BPRP!
Blackpenredpen I am genuinely interested in being an expert in algebra and number theory. I am currently in grade 12(A2). Could you please suggest me some books? I really enjoy your videos and they have really influenced me. It would be like a dream come true to take suggestion for brainy maths book(the book with sufficient examples and questions) from a great teacher like you.
I've literally been working on this exact problem, except it was abs( ln(x )
Just a question in my mind, since the solution is complex (as expected), is there any periodicity in the stated solution?
First multiply both sides by x and the take w lambert on both sides to get e^x=x. Next multiply both sides by e^-x and -1. Take w again and multiply both sides by -1 to get answer: x=-lambert w(-1)
Thank you sir!
0:57 does it work if the function is monotonic in general too?
Is it possible that we use Euler equation for getting the solution sooner ?
I have that same bob-omb plushie haha
where'd you get it?
@@bluepeacemaker I remember like getting it from an arcade I think
Thanks thanks a lot for cutting off that checking bush
A good question be find the minimum distance between e^x and ln(x)
Hint-line x=y
Sorry if I am in the wrong place but I vaguely remember from my school days that
X^y = exp(y*Log(x)) with x,y real numbers
I am probably completely wrong, but if I am right, I fail to see why this is the generalization to R of: "to the power of". What does "powering" have to do with Log end exp?
Can you make video how to arrange matrix 4x4 to diagonally dominant ?
I want to solve equation with gaus seidel iteraton
أفضل أستاذ في الرياضيات هو الأستاذ نوردين ،،،، الجزائر ،،،، 🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿
Hey dude! Can you do limit of (3/2)^n*((sqrt(3^(2)+5n*2^n)-sqrt(3^(2n)+4))/n) as n goes to infinity? I think it should be fun and I would like to see your approach. Thx and have fun
As someone pointed out, solving e^(e^x)=x ( equivalent to your equation) is not the same as solving e^x=x. In general, for any function f, f(f(x)) =x can have more solutions than f(x)=x: the fixed points of f are a subset of the fixed points of f composed with itself.
I am new for this kind of maths
Why can’t we take natural log on the both side when e^x=x?🤔
Is it no way to get the complex number?
It'll give you lnx = x which is also unsolvable in the reals since lnx is smaller than x
Stirner's Ghost I see, thank you
My solution:
> e^(x) = Ln(x)
By multiplying both sides by 'x' ,
> xe^(x) = x*Ln(x)
By taking the productlog of sides:
> x = Ln(x)
> e^(x) = x
> e = x^(1/x)
> e = x^(1/e)^(Ln(x))
> e = e^[(Ln(x))[(e)^(-Ln(x))]]
By taking the natural logs,
> 1 = Ln(x)e^(-Ln(x))
By multiplying both sides by '-1' ,
> -1 = -Ln(x)e^(-Ln(x))
By taking the productlogs;
> W(-1) = -Ln(x)
> -W(-1) = Ln(x)
Therefore;
x = e^(-W(-1)) = -W(-1) = Ln(-W(-1))
Is there a video on Lambert w function?
The cool head solution mann !
i don't even need to solve this, i'm just watching cause i like your videos
excellent videos I can propose one of analytical geometry to know more about this topic with your help
The microphone looks like a Mate from argentina, also greetings from here!!
How about a video on 100 logarithms, especially challenging ones?
Does the "tower of e" says that e^x = ln(x) if lim x-> inf ?
Someone may have already mentioned this? One can manipulate the equation e^(e^x)= x by multiplying both sides by e^x. Then applying the Lambert W gives e^x = x. The rest of the solution is as you showed. That eliminates the need to explain e^e^e^e^.....
i am new to this channel. can somebody explain why he holds the bomb from mario in his hand?
can you do a video on how to evaluate natural log (ln) without a calculator
how about just differentiate xe^-x and find out that it has maximum at 1 and use 11/e= f(1) ?
That was a convoluted way to use the property that solving f(x) = f^-1(x) is the same as solving f(x) = x.
e^x = ln(x)
e^e^x = x
substituting the equation for x you get an infinite power tower of e^e^e^..., so you can just write e^x = x
bring to one side: xe^(-x) = 1
-xe^(-x) = -1
using the lambert-W function: W(-xe^(-x)) = W(-1) = -x
x = -W(-1)
according to wolfram alpha its approximately equal to:
0.318 - 1.337i