e^x meets ln(x)
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- čas přidán 9. 05. 2021
- We will make b^x and log_b(x) tangent to each other here: • the famous equation b^...
Exponential function and logarithmic function! We will discuss this hard calculus problem on how to move e^x horizontally so that it will finally meet ln(x). Check out this video for solving e^x=ln(x) in the complex world: • Is e^x=ln(x) solvable?
Here’s a video on finding the minimum distance between e^x and ln(x) • Minimum distance betwe...
A complex solution for e^x=ln(x) • Is e^x=ln(x) solvable?
Lambert W function explained: • Lambert W Function (do...
Omega Constant, i.e. W(1): • Newton's method and Om...
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blackpenredpen
Woah, I didn't even expect an answer from such a big CZcamsr, but you even made a video about my question. That's so cool. I have been troubled by this problem since the day we got taught exponential equations in high school. I've always wondered how to solve those when both exp() and ln() functions show up in the equation .Thank you so much! -Mark
Here’s the man!!
I like this question a lot! So thanks for that!!
Elegant question yielded that elegant solution !
@@blackpenredpen i have almost the same question:
When a^x=log_a(x) and has only 1 solution.
@@Th3AnT0in3 a=e^(1/e). if you want me to make a video about it i can do that
@@SPVLaboratories omg you got it, what are the calculous i need to do to find this answer ?
I tried it a few years ago and i failed but idk if i can solved it now because i'm better at maths
(Sorry for "frenglish" btw 😋)
Note to self: Do not watch bprp before morning coffee. Brain now hurts.
😆
@@blackpenredpen this video is freaking good i like you so much
yeah dude now it hurts ssoo much......aaaaaaahhhhhhhhhhh
I dont think Ive ever seen you enjoying yourself quite this much. The look of joy on your face when you got them to touch brought a smile to my face.
Haha its funny to see the Lambert W Function pop almost in every bprp video!
Would love to see a lecture on how the W function is derived and how it works
It's more of a place holder than anything. You need to use Newton's method to solve it.
He has an explanation on how it works on another video
Please see description for the video 😃
@@blackpenredpen Thanks alot! I searched on YT for it but it didn't show up somehow :)
My fault for not looking at the description
BPRP: It makes no sense how big this number is
Expects something in scientific notation
BPRP: 2.33
Confused pikachu face
Long live (in our hearts, at least) to the soviet union, the first great attempt to leave behind our pre-history!
Well, he did draw the solution on the board at the start
Such a joy to see you solving and explaining them with clarity, great job mate!
We will make b^x and log_b(x) tangent to each other here: czcams.com/video/uMfOsKWryS4/video.html
cool fact i just realised: applying a translation to exp is exactly like scaling it in the y direction, since e^(x-a) = e^x/e^a . Just like exponential transform addition into multiplication, it transform translations into homothety
That was sick. Great work bprp
Amazing video as always! Great thinking by mark. Loved his idea.
Watching this video before surgery, your math always brings me happiness and joy :)
Instead of shifting the graph, how about changing the base instead? Like what number "a" such that the graph a^x and log(a,x) touches?
Great question! It happens for a=e^(1/e).
@@SabyasachiGhosh1618 so the 'e'th root of e
Where we have a flat line and something imaginary?
He did it today 😄
I like how it’s a simple question with a cool answer.
I always wanted to figure this out, but didn't really know how. Thanks for doing this!
The final question for our univ entrance exam in South Korea 2 years ago was actually very similar to this question! It's very interesting that you happen to show this problem in your video today.
wow. nice video. its nice solving challenging calculus questions of this sort and that is what i love doing on my...... thanks for checking t out
Loved this innovative manipulation 🧡
This is cool,that helps everyone who wants the subject math.
Nice! I did not know about the Lambert Function until I watched your videos! You're amazing! 🤩
Really cool topic. I had made a video on the intersection of y=e^x and y=kx but my wording was incorrect. I asked for the k value for only one solution to e^x=kx but my intention was "What is the k value if the graphs are tangent?" which has the same idea.
One of the few channels which makes maths fun.
great
@Virat Kohli you can easily tell that from my name.
BTW what's your name?
Agreed
Congratulations 🎉🎉 sir for 700k , soon 1m
1/W(1) + W(1) is equivalent but a bit cleaner in my opinion (you can show from W(x)e^(W(x)) = x that e^(W(x)) = x/W(x) and ln(W(x)) = ln(x) - W(x), plug in x = 1 to obtain e^(W(1)) = 1/W(1) and -ln(W(1)) = W(1).)
It's ALL relative !
I can’t believe I understood that. I’m not native English and didn’t have integrals in school yet. Your explanations are amazing❤️
Very beautiful Steve sir keep uploading stuff like that Sir 😀
OMG I LOVE HIS SHIRT
This is why I love using the Desmos graphing Calculator
Such a nice answer
Finally the Crossover we needed
i love that you're using version 5 of geogebra.
Amazing video
a beautiful exercise.
Now we do e^x= ln(x+a)!
Actually, no need. ln(x+2.33) =e^x as we might expect and BPRP alluded to in the beginning. If you move both curves than there are infinitely many solutions right? Cool that x = 2.33 involves "omega" though 🤷♂️
beautiful.
That was actually cool 😎
Thanks!
Sir, how to solve this series problem:
5,7,17,55,225,1131, x , 47559
Find the value of x??? Sir, make the video on this topic please.
I have also try but I didn't get answer.
Can you tell me the answer?
@@abhid5300 Ok , But first of all blackpenredpen give answer.
@@MATHSEXPLORER1 Yes, I have try but it doesn't make any formula, please tell any hint.
Any value of x fits.
7 = 1*5+2
17 = 2*7+3
55 = 3*17+4
225 = 4*55+5
1131 = 5*225+6
6793 = 6*1131+7
47559 = 7*6793+8
Really cool
Nice one!
Amazing. Could you plz also give some STEP 3 questions a go. They are quite a lot more difficult and are more beautiful than the STEP 2 question than you attempted as well. There are some very beautiful one such as proving the irrationality of e etc.
Finally, a real mathematic video after a long gap
Really Cool !
And you have that the value of the functions in this point (χ=1/Ω) it's Ω and its derivates 1/Ω. Really Nice problem
I reminds me a question i had in a math exam in High school, where we had to find the smallest 'a' such that ax^2 = ln (x) only have 1 real solution . It took me a while to figure it out tbh
Top ten greatest love stories 😂
😆
Considering this was relatively easy, i wondered if it was actually solveable in a general case for moving in two directions (x and y), so:
e^(x-a) = ln(x) + b
Obviously this would generate a whole set of solutions itself, and ideally one could try to look for the minimum of this set in terms of "distance moved", i.e. minimum of c = a^2 + b^2, and i think this should give one (or mulitple?) Solutions.
Turns out i am shit at math though so i got stuck in the process of getting a function nice enough to differentiate in terms of c.
Just leaving this here in case any smart person comes around to this :)
@Henry 1 yeah this is exactly right. you can do some insane Lagrange multiplier/Lambert-W manipulations to get the same thing but this is a good intuitive way to look at it
I asked myself the same question. The solution is f(x)=e^(x-1)-1 and it is tangent to ln(x) in P=(1,0). I started in the same way bprp did:
e^(x-a)-b=ln(x)
same tangent so:
e^(x-a)=1/x
now, instead of explicating x, using the Lambert W function, we explicate a and b:
b=e^(x-a)-ln(x)=1/x-ln(x)
a=ln(x)+x
so we have
a^2+b^2=(ln(x)+x)^2+(1/x-ln(x))^2
deriving this function and setting the derivate to 0, we have a long equation that has only one real solution that is x=1. So we have:
a=ln(1)+1=> a=1
b=1/1-ln(1)=> b=1
I see (@Henry 1, @92ivca) should have tried a few easy cases first lol. Would be cool to see someone tackle an actual analytical solution of this, but i think the math might actually melt my braincells.
@@erik9671 the math isn't that hard, I edited my previous answer with a solution, but it doesn't shows all steps, because I ended up solving a very long equation
@@92ivca Oh i see, thats really not thaaaat long, thought thanks for the edit :) (I am an engineer and we generally take sin(x) = x as an approximation regarless of the angle, so math stuff that is pretty easy can sometimes fuck me up pretty badly)
I knew bprp was talented, but now inverse-function matchmaker?!! Wow!!! Next, he'll be performing the wedding ceremony for the sine and cosine functions.
We also have a = W(1) + 1/W(1) which I think is beautiful than the formula given in this video (but as I said it is just my way to think).
Good video btw 👏😉
Lover your channel. An idea for a similar problem that could be interesting to see your solution for:
Finding base b so that y=b^x tangents the related b-base logarithm y=logb(x) in one point.
Thanks! I actually recorded that video a few days ago. Here’s my Chinese version
czcams.com/video/GDS6VWgcUXw/video.html and my English version will come out this week.
@@blackpenredpen Great! Looking forward too it:)
Cảm ơn ad rất nhiều thank you very much i am from vietnam
Not going to lie, I understood everything until you slapped W in there.
A much more accessible problem (and I think more fun) is the following: Suppose we wish to find a base "a" such that a^x = log_base_a(x) at only one point. In other words, we want to find the exponential and log base that makes these two functions just touch one another at one tangent point. Using first-year calculus only, you will find that a = e^(1/e). COOL!
Paused and worked out on my own, and got it right a completely different way! Here’s what I did.
I set a = z to make it a 3D surface, y=e^(x-z), then solved for z=f(x,y). I then changed y=lnx to be 0=lnx - y as a level curve constraint. Then I used Lagrange multiplier and gradients, and solved for λ. Unfortunately, there was no solution I could find by hand to lnλ=(1/λ) so I converted to (1/λ)e^(1/λ)=1 and used Lambert W to solve. After plugging in the solution for λ, which was 1/W(1), I found the shared normal vector of , intersecting at (1/W(1),W(1)) or about (1.763, 0.567). I then plugged those x,y into the z=f(x,y) function to get about 2.33 for the “z” distance, which is the “a” distance.
Using a metal sledgehammer to break a piece of butter@IonRuby
aye im happy i did it all mentally
Hi, we could move both the curves so that they touch the line x=y since ln and exp are inverses of each other. That could be a neat solution as well.
1M incoming 😀
A mathematician's version of a meet-cute right here.
Interesting. Was wondering what if you move e^x to the right AND ln(x) upwards until they meet. In other words, for what shift a do e^(x-a) and ITS INVERSE ln(x)+a meet? The answer is simply for a=1, at x=1
I can get to W(e^a) = e^(1/W(e^a) - a). I can see that a = 1 solves it, but can you solve it algebraically?
@@miruten4628 No... I tried for a while to no avail, got that solution just by inspection. In fact, Mathematica gave up on {e^(x-a)==log(x)+a, e^(x-a)==1/x} with both Solve and NSolve
If f(x) and f^(-1)(x) meet at some point, this point should be on y=x , isn't it? I'm not sure whether it always hold, but in this case, it allows problem to be solved easily. e^(x-a) = x && e^(x-a) = 1 = > x=1 => e^(1-a)=1 => e^(1-a) = e^(0) = > 1-a = 0 => a= 1
@@miruten4628 you are thinking way too complicated. the functions are symmetric to y=x, so they actually have to touch that line at their meeting point. you get the equation ln(x)+a = x with derivative 1/x = 1, which solves easily to x=a=1
Its so intriguing the type of numbers that appear when we ask ourselves these stuff.
Is w(1) transcendental?
Nice 👍
Love bprp 💘
You used to make a bunch of videos on integrals. I wonder whether you could integrate sqrt(ax^2 + bx +c), maybe call it the King Euler-substitution.
From the relation e^W(1) = 1/W(1), you can also show that W(1) = -ln(W(1)), so the final answer can actually be written more simply as: a = W(1) + 1/W(1)
What are the mechanics of how a computer calculates the LambertW function?
I solved it differently: You shift along y=const. So I tried to find a horizontal which intersects the two graphs at points with a common derivative. For that I needed the derivatives with respect to y. So I solved the functions y(x) for x and differentiated for y resulting in 1/y and e^y. Setting them equal you immidiately find y=W(1). Now you simply plug that into the x(y) and immidiately get a.
Or more elegant: Because the problem is symmetric under an y-x-switch aka when you mirror along y=x, nothing changes, you can instead ask yourself, by how much the ln needs to be shifted up. This way you skip the solving-for-x-step and the confusion it brings:
So e^x=lnx +a -> d/dx -> e^x=1/x => x=W(1) => a=e^W(1) - lnW(1)
You can also try to meet them by lifting up ln(x): eˣ=ln(x)+a
Hey Blackpenredpen!
I’ve stumbled upon a problem which I don’t really know how to solve... how do you integrate the x-th root of x?
Finally someone decided to update math
Because of W's weird relationship with e, a can also be expressed as 2cosh(W(1))
Cool 😃
I did a similar solution that eventually significantly deviates:
Instead of starting with e^(x-a) = ..., I stated that "As the derivative of e^x is itself, it can only tangent where ln(x) intersects with its derivative, thus we must find out where ln(x) and its derivative meet", leading to the same equation.
There, I used e^() instead of ln() to eventually get to x^-1 e^(x^-1) = 1 which also ends up with x = 1 / W(1).
Here, the steps change significantly. I instead calculated the y value of the intersection. This is quite simple, as I just inserted the previous x value into 1 / x:
y = 1 / x
y = 1/ (1 / W(1))
y = W(1)
I then determined where e^x meets that y value. This required the identity that ln( W( x ) ) = ln( x ) - W( x )
e^x = W(1)
x = ln(W(1))
x = ln(1) - W(1)
x = - W(1)
Finally, I determined the value a by using the difference between the two previous x values.
a = W(1)^-1 - ( - W(1) )
a = W(1)^-1 + W( 1 )
While the solution looks different, it equals the same value as the one in the video.
You should do a part II problem where the function f(x) = ke^x touches ln(x)
Any chance you can make a calc series on a playlist?
Here’s a similar Challenge:
A circle with center (2,6) and a radius of r is tangent to the parabola y=-2(x-6)^2 + 6 at one point. Find the value of r
That's easy
@@ijemand5672 ok
Is it ≈3.43905
@@tanishqrulania9902 👍
Explanation please?
Since the answer is approximate, is there any issues with precision? In other words, if the two functions must intersect at exactly one point, what guarantee do we have that ~2.33 is sufficient to meet that criteria?
When even a mathemarical formula finally meets someone but you don't
engineer's watching: "wow so the answer is e! I did not expect that!"
I prefer e^(x-a)-a. It’s not much of a challenge but they are parallel at x=0,1
It’s so much nicer geometrically.
If we're talking simmetry, I preffer e^x - 1 and ln(x+1) it touches on (0,0) and is symmetric on the y=x line
@@aronmaciel 😂 nice one
if it had a solution it would be like 2 parallel lines meeting together...ROFL
At this point I feel like this is Lambert W function the channel
Is this a new transcendental number? Can that number do anything else? Does that number describe any other relationships?
讓我想到，想請問曹老師，如果e^x是以（0, 1)作為原點，進行旋轉。lnX不動。那麼，兩個曲線相交於僅一點的時候，會是哪兩個點呢？（逆時針、順時鐘 應該各一點吧？）或是有辦法求兩點距離嗎？感謝您～
Hey Stanley, 這題別人也有問過, 我得好好想想!
When you set e^{x-a} equal to \ln(x), how can you be sure that there is only one solution? e.g., if a were to be larger than the value we found there should be 2 values of x s.t. these functions give the same image at those values.
IF: a = ???
then: e ^ (x - a) = (e ^ x) - a
(aka irrelevance of parenthesis)
I solved what seems like a variation of the original question: how far to “lower” e^x so it can meet ln(x)??
After achieving the solution, I was forced into a brief pause. Then I had to “duh!” myself. And then, after another brief pause, I had to “whoa, cool!” myself.
Still not sure if I was smarter before or after I solved the “other” case.
interesting thought. move the curve to meet at a point. But do they really converge? In terms of a curve from a two-dimensional graph, yes they are contiguous. But what happens when we look at this equation in a three-dimensional graph. Is it just an optical illusion because we use an X / Y graph?
This is like me talking and meeting myself in the mirror
I can't find the version of the black board cheat sheet you have on your wall, I only see the derivative one
I have a follow-up question from this same equation between e^x and ln(x). What if instead we rotate e^x clockwise, what is the angle needed to rotate so that e^x will touch ln(x)?
You didn't fully simplify a in the video as you could have written it as
a = e^w(1)+ln(1/w(1))
a = e^w(1)+ln(e^w(1))
a= e^w(1)+w(1)
Or 1/W(1) + W(1) if you want. But his way of solving is the thing that matters. :)
Next step: since the translation of e^x shown is only in x direction, find another translation (in both x and y direction) that minimize the translation length.
(Hope you understand what I mean... Sorry for my bad english)
SPOILER:
Don't know if my math is correct but I found a really satisfying solution:
e^(x-1)-1
that is tangent to ln(x) in P=(1;0)
No Lambert W function needed to found this
@@92ivca Nice, how did you find that?
@@spaghettiking653 same start, but with "a" and "b":
e^(x-a)-b=ln(x)
e^(x-a)=1/x
Now we can explicate a and b (instead of x)
a=ln(x)+x
b=1/x -ln(x)
We need to minimizing this:
a^2+b^2=(ln(x)+x)^2+(1/x -ln(x))^2
Searching the zeros of the derivate of the function I found only one real solution, x=1
Excellent and elegant request !
@@92ivca Thanks, good work !
Other than shifting e^x on the X axis we should find also the solution if we shift it on Y, so we have e^x - a rather than e^(x-a) always equal to lnx ofc
what about determining the shortest distance between the two curves and what are the two points of the straight segment?
Or by inspection you could do a = e so that they meet at the point (e, 1)
its actually much easier, if you shift both functions. since the graphs are symmetric with respect to the graph of y=x, you can just shift them, so that they touch the line. that way you get a picture that is much more symmetric.
and the equations are ln(x)+a = x, 1/x = 1, which easily solves for x=a=1
so you get e^(x-1) and ln(x)+1
The inverse function of y = e^(x-1) is y = ln(x)+1, so they meet on the line y = x at point (1,1) where their derivatives also have the same value 1, but of course this isn't as cool as shifting only y = e^x to the right and dealing with the Lambert W function ordeal it causes.
I have no idea what W(1) is, gonna check it out now
My first instinct was to equate e^(x-a) and ln(x). That would give me a solution for x which can be 0, 1 or 2 values depending on the value of a. Then I would try to find the a that makes the solution be one value (like when we try to make a discriminant 0). Would that work?
could you solve by sucessive substitution if it is a convergent problem?
am I right ,Sir ln(X) +2and e^x also touch together and ln(X) +3 and e^x are intersect at two point.
Everybody gangsta till bprp brings out a blue pen
very cool! also, maybe I'm missing something, but where x=e^(W(1)) and a=e^(W(1)) - ln(W(1)), doesn't x-a just = ln(W)1? so e^(x-a) = e^ln(W(1)), which would just equal W(1)?
I guess my question is, is this whole equation equivalent to W(1) = ln (x)?
You can simplify that equation since exp(W(1)) = 1/W(1) and ln(W(1)) = -W(1), so a = W(1) + 1/W(1)
Bob Ross would be proud of you! :-)