Is e^x=ln(x) solvable?
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- Опубликовано: 27 авг 2024
- 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: ruclips.net/video/uMfOsKWryS4/видео.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?
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: ruclips.net/video/53lBKCBrENY/видео.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
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
I was going to ask this exact question since:
exp(x) = ln(x)
x exp(x) = x ln(x) = exp(ln(x))ln(x)
W[x exp(x)] = W[ln(x)exp(ln(x))]
x = ln(x)
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 ∪ {∞})
Simple math:
BLACK PEN
RED PEN
Complex math:
BLACK PEN
RED PEN
BLUE PEN
.
.
.
even more complex math:
+Purple Pen
@@Killer_Queen_310 quaternion
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
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)
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
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?
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.
Reals: Nope..no solution to this thing..
Complex numbers: Hold my “i”s
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
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
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
i stared at the thumbnail for five minutes thinking before clicking. this strategy works
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
Test: solve e^x=ln(x).
me: "eeeeeeeeeeee...."
black pen red pen, and occasionally, blue pen
I loved the 🐧,the song and not to mention "we are not doing real mathematics"✌🏻😌
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.
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
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.
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.
Since e^x and ln(x) are inverse functions, as you drew. The only places where they can intersect is when the mirroring in the line y=x is on the same point. That means that the output of e^x=x=ln(x).
This is a faster way to arrive at e^x = x. This always works for inverse functions.
I miss the intro from 2017: " Black Pen Red Pen yaaaay!" Like if you agree
@Tropical_Papi It was cringe.
@@adityakamat9856 no
first time heard "lambert W function".
Before: Black pen red pen.
After: Here comes blue pen.
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)
This is one of my favorite channels!
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
Love from India . I greatly admire your maths skills and teaching .
Jai Bharat!!!! 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
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
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)
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?
As we know ln x < X < e^x for every real number x>0. So obviously this equation has no solution in the real field. So you have to find the solution in the complex field which unsurprisingly has a solution to almost every unsolvable equation in IR. Except equations like (1/z =0 for example).
Having to increase the amount of e’s in the power tower all the way to infinity to be able to reduce it to a single one 🤯
My reasoning was that lnx is the inverse function of e^x,so ,since doing the inverse basically means doing the simmetry of the function along the line y=x,solving e^x=ln(x) means finding the points of e^x that are on the line y=x
Note: if e^x = ln x,
then x e^x = x ln x,
ànd you get x = W(x e^x) = W(x ln x) = ln x,
faster than the infinite e^e^e...^x
But it is only one lf the solutions. There are two solutions to W(x ln x) or W(x e^x).
And more than two in complex space.
A propos, why does not your answer also accept the conjugate ? Typically if ln x = a + bi, then ln(x') = a - bi So x' is an answer too.
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
e^x=ln(x) ==> e^(e^x)=x ==> (e^x)*(e^(e^x)) = (e^x)*x ==> apply lambert ==> e^x = x ==> divide by -e^x, you get -1= -x*(e^-x) ==> apply lambert x = -W_n(-1)
A good question be find the minimum distance between e^x and ln(x)
Hint-line x=y
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^.....
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
"pause the video, and think, about, this."
"... and today we have this guy. ok so-"
😂
You know it gets complex when blackpenredpen includes a blue pen.
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.
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))
What I was looking for, a video for the complex natural log. Thank you so much B&Rp!
Okay, there is a more concrete way to prove the equation comes down to e^x = x
If f is strictly increasing, then the equation : f(x) = inverse (f(x)) (1), has the same solutions as the equation f(x) = x. If someone asks I am gonna provide a proof of that
But "strictly increasing" is not a defined concept in complex space. So your result is only valid for real x and f.
A funny thing to notice here is that e^x = ln(x) = x is exactly the first term of the Taylor expansion of ln(x) at the point where they should meet (x=1)
ln(1) ≠ e^(1)
@@privateaccount4460 The first term of the Taylor series for each is 1.
i don't understand what you mean
Best microphone!
Thanks 😆
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.
أفضل أستاذ في الرياضيات هو الأستاذ نوردين ،،،، الجزائر ،،،، 🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿🇩🇿
Holy crap ! I actually thought and tried to solve the equation and got the correct answer . Never thought could do it
If you find the Taylor expansion of lnx and e^x then group all the like powers of x on one side. then by the fundamental theorem of algebra you have infinitely many solutions that are all complex as the two functions never intersect in the real numbers. You also have conjugates as by the real coefficients in the Taylor polynomial. I guess this isn’t that helpful as it doesn’t find the solution but it shows what they look like.
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)
by 1:58 i saw it- lambert W! so: x = e^e^x, multiply by e^x, and we get x e^x = e^x e^(e^x). W both sides and you get x = e^x. this also doesn't have a real solution, but no matter! divide by e^x and negate both sides: -xe^-x = -1, so -W(-1) = x
Just learned about log and ln in class, this blew my mind
Watching from South Africa. You made me love math man! Keep feeding me
Hey man love your t shirt
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!!!! 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
It's very interesting! I don't understand English very well(?), but, the process you write on the whiteboard is very easy to understand!
e^^infinity would have been the perfect oppurtunity to make practical use of tetration.
That was a convoluted way to use the property that solving f(x) = f^-1(x) is the same as solving f(x) = x.
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)
W(.) has several values.
Actually, there ARE solutions where x is not ln x. That is, where y = e^x = ln x , yet y =/= x
To find them, establish x = a + bi = re^it
We seek z(x)= e^x - ln x = u + vi = 0.
this is: z(a,b) = (u,v)
Write the Jacobian dz/dx :
[ e^a cos b - a/R2 ; -e^a sin b - b/R2 ]
[ e^a sin b + b/Ta^2 ; e^a cosb - 1/Ta ]
where R is a^2+b^2 and T is 1+t^2, t being the angle arctg(b/a).
Then by Newton-Raphson, iterate
Dx = Jinv • (-z)
Depending on the seed value, you will find solutions of the first kind (x=lnx) : 0.318 +- 1.337 i,
or the second kind (x=/=lnx) : 1.9 +- 1.4 i .
Why is the infinity argument incorrect? because there is an infinity of DOUBLE exp. So when you take out ONE, the remaining sequence is odd, hence not provably equal to x.
You’re gonna need more pen colors if you keep doing problems like this.
Love the + C shirt. Only true math geeks get that one. 🤣
‘True math geeks’ aka anybody who’s taken an introductory calculus course.
@@randomblueguy was gonna say that as well lol
What's that
+C is the constant that must be added when calculating the anti-derivative, right?
@Colinho Abrahamovich
yes
Are we able to put e^e^e^… =x even it doesn’t converge? Mine is the following. Is it valid or not?
e^x = ln x --> e^(e^x) = x --> (e^x)e^(e^x) = x(e^x) --> Apply W --> e^x = x --> -1 = (-x)e^(-x) --> Apply W --> x=-W(-1)
I have that same bob-omb plushie haha
where'd you get it?
@@bluepeacemaker I remember like getting it from an arcade I think
Lambert W function exists:
*blackpenredpen
: I can milk you*
wait. does this mean e^x=lnx e^x=x ?
Yes, he mentions that if f(x) = f^-1(x) its the same as f(x) = x. This is because taking the inverse of a function, its the same as reflecting it along the line y = x. This means that a function and its inverse will always intersect at a point on the line y = x. So finding the intersection of f(x) with y = x is the same as finding the intersection with the inverse function. However, since neither e^x nor ln(x) intersect with y = x this problem has no real solutions.
We could go to e^x = x in another way:
Replace e^x = a => e = a^(1/x)
e^e^x = x
(a^(1/x))^a = x
a^(a/x) = x
a^a = x^x
a = x
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.
Black pen red pen you are the best at what you do
I really like how him prove e^x = e^(e^x)
you are really good at maths
I was never thinking there are fixed points for exp() (except for transfinite numbers). Nice to know.
Bruuuuuh I just solved that yesterday before I found this video. I wonder if we'll use the same method.
EDIT : I'll put my method here just I case :
e^x = x
=> e^e^x = x
=> e^x = ln x
So solutions for e^x = x work for e^x = ln x
e^x = x
e^-x = 1 / x
xe^-x = 1
-xe^-x = -1
-x = W(-1)
x = -W(-1)
EDIT : Just watched a bit more of the video and apparently we have the same thing going
"hello, let's do some math for fun"
ctrl+w
Interestingly, Mathematica 11 can't find the -W(-1) solution directly, even though it can verify it (FullSimplify[Exp[x] == Log[x] /. x -> -ProductLog[-1]] returns True)
b/root(n,b)=x
given x and b find n
beacuse the mirror line of the inverse function is at y=x you can say when f(x)=f^-1(x), f(x)=x
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 would really love to see how to compute that lamber w function value at the end
i love the sound effects
Ah... Finally, come back with another video. ❤️❤️
Another way to approach it is to imagine x=r*cis(theta), so that e^x=e^[r*cos(th)]*cis[r*sin(th)] and ln(x)=ln(r) +i (th)... But that's a bit complicated for me...
Since x=ln(x), it follows that r*cis(th)=ln(r) + i* ln(th), so r*cos(th)= ln(r) and r*sin(th)=ln(th)...
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.
One of the solutions to this is infinity because e^inf=inf and ln(inf)=inf
Please solve this question n^3/3^n sum from 1 to infinity
you could've simple multiplied by e^x in both sides in x=e^e^x
and then applied lambert W function in both sides you would've got x=e^x instantly.
See above.
W has several branches. This result is true but not unique.
I set x = a + bi, assumed a = 0, moved to the complex plane, and came back with x = (n pi i)/ 2. I’ll assume that’s correct on another branch of the Lambert function.
i set x to W(x)
5:05 i thought the fish is going to be summoned...
It's weird that the functions don't touch each other in XY graph, but the complex solution still has a real part different than zero
I'm confused about it,,,
Please help me spotting my mistake,,
e^x=lnx
e^x=1/x (differentiate both side)
xe^x=1
W(xe^x)=W(1)
x=W(1)
But I calculated it it shows e^x=0.5671
And lnx=-0.5671
3:25 PAGING DR PEYEM! PAGING DR PEYEM! YOU ARE NEEDED HERE!!!!!
how about just differentiate xe^-x and find out that it has maximum at 1 and use 11/e= f(1) ?