e^x meets ln(x)

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

[@Astri.electronics]

Woah, I didn't even expect an answer from such a big KZbinr, 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

[@blackpenredpen]

Here’s the man!! I like this question a lot! So thanks for that!!

[@deedewald1707]

Elegant question yielded that elegant solution !

[@Th3AnT0in3]

@@blackpenredpen i have almost the same question: When a^x=log_a(x) and has only 1 solution.

[@SPVLaboratories]

@@Th3AnT0in3 a=e^(1/e). if you want me to make a video about it i can do that

[@Th3AnT0in3]

@@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 😋)

[@MathAdam]

Note to self: Do not watch bprp before morning coffee. Brain now hurts.

[@blackpenredpen]

😆

[@aashsyed1277]

@@blackpenredpen this video is freaking good i like you so much

[@aryanabhilesh11]

yeah dude now it hurts ssoo much......aaaaaaahhhhhhhhhhh

[@wjshood]

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.

[@Kdd160]

Haha its funny to see the Lambert W Function pop almost in every bprp video!

[@benedictspinoza1025]

BPRP: It makes no sense how big this number is Expects something in scientific notation BPRP: 2.33 Confused pikachu face

[@gabrielnettoferreira8452]

Long live (in our hearts, at least) to the soviet union, the first great attempt to leave behind our pre-history!

[@ieatgarbage8771]

Well, he did draw the solution on the board at the start

[@8-P]

Would love to see a lecture on how the W function is derived and how it works

[@damonpalovaara4211]

It's more of a place holder than anything. You need to use Newton's method to solve it.

[@justinpark939]

He has an explanation on how it works on another video

[@blackpenredpen]

Please see description for the video 😃

[@8-P]

@@blackpenredpen Thanks alot! I searched on YT for it but it didn't show up somehow :) My fault for not looking at the description

[@blackpenredpen]

We will make b^x and log_b(x) tangent to each other here: kzbin.info/www/bejne/q37JgKaBjNesiZY

[@SpaceWithSam]

Such a joy to see you solving and explaining them with clarity, great job mate!

[@Joffrerap]

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

[@3ckitani]

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?

[@SabyasachiGhosh1618]

Great question! It happens for a=e^(1/e).

[@ManjotSingh-sf2ri]

@@SabyasachiGhosh1618 so the 'e'th root of e

[@loganroman5306]

Where we have a flat line and something imaginary?

[@92ivca]

He did it today 😄

[@Melkboer38]

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).)

[@deedewald1707]

It's ALL relative !

[@andreiion6395]

Watching this video before surgery, your math always brings me happiness and joy :)

[@tomileevasico5741]

This is cool,that helps everyone who wants the subject math.

[@VikashKumar-pj6bs]

One of the few channels which makes maths fun.

[@mathsandsciencechannel]

great

[@VikashKumar-pj6bs]

@Virat Kohli you can easily tell that from my name. BTW what's your name?

[@abcd1234___]

Agreed

[@sueyibaslanli3519]

Finally, a real mathematic video after a long gap

[@MATHSEXPLORER1]

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.

[@abhid5300]

I have also try but I didn't get answer. Can you tell me the answer?

[@MATHSEXPLORER1]

@@abhid5300 Ok , But first of all blackpenredpen give answer.

[@mohansingh3750]

@@MATHSEXPLORER1 Yes, I have try but it doesn't make any formula, please tell any hint.

[@gavasiarobinssson5108]

Any value of x fits.

[@debblez]

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

[@erik9671]

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 :)

[@SPVLaboratories]

@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

[@92ivca]

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

[@erik9671]

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.

[@92ivca]

@@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

[@erik9671]

@@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)

[@Twitledum9]

Now we do e^x= ln(x+a)!

[@Twitledum9]

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 🤷‍♂️

[@jayska5802]

That was sick. Great work bprp

[@Asterisk_766]

Finally the Crossover we needed

[@SyberMath]

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.

[@icantseethe7680]

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

[@ijemand5672]

That's easy

[@antonhelsgaun]

@@ijemand5672 ok

[@tanishqrulania9902]

Is it ≈3.43905

[@icantseethe7680]

@@tanishqrulania9902 👍

[@zzztriplezzz5264]

Explanation please?

[@shadmanhasan4205]

This is why I love using the Desmos graphing Calculator

[@92ivca]

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)

[@92ivca]

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

[@spaghettiking653]

@@92ivca Nice, how did you find that?

[@92ivca]

@@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

[@deedewald1707]

Excellent and elegant request !

[@spaghettiking653]

@@92ivca Thanks, good work !

[@hassanniaz7583]

Amazing video as always! Great thinking by mark. Loved his idea.

[@mathsandsciencechannel]

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

[@samuelromero1763]

I like how it’s a simple question with a cool answer.

[@michaeleiseman4099]

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!

[@Rolancito]

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

[@miruten4628]

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?

[@Rolancito]

@@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

[@penguinpenguin-zm2mr]

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

[@nasekiller]

@@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

[@toonoobie]

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)

[@NoName-eh8fz]

Or 1/W(1) + W(1) if you want. But his way of solving is the thing that matters. :)

[@Mark16v15]

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.

[@Oliver-wv4bd]

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)

[@carultch]

What are the mechanics of how a computer calculates the LambertW function?

[@polyhistorphilomath]

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.

[@aronmaciel]

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

[@allenjonesstyles6112]

@@aronmaciel 😂 nice one

[@madnessJATIN]

Congratulations 🎉🎉 sir for 700k , soon 1m

[@reussunased5108]

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

[@ThAlEdison]

Because of W's weird relationship with e, a can also be expressed as 2cosh(W(1))

[@ActionJaxonH]

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.

[@MrFeast-l1d]

Using a metal sledgehammer to break a piece of butter@IonRuby

[@dhoom-z7221]

Top ten greatest love stories 😂

[@blackpenredpen]

😆

[@renyxadarox]

You can also try to meet them by lifting up ln(x): eˣ=ln(x)+a

[@davidb2885]

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)

[@Ze_eT]

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.

[@valemontgomery9401]

I always wanted to figure this out, but didn't really know how. Thanks for doing this!

[@dylanl.3366]

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.

[@Bangaudaala]

2:40 HE CANT KEEP GETTING AWAY WITH THIIS😭

[@uaswitch]

A mathematician's version of a meet-cute right here.

[@alexanderlea2293]

engineer's watching: "wow so the answer is e! I did not expect that!"

[@alessandronitti6941]

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

[@nishantkumartiwari1202]

Calculating (-1/2)! by a method adopted by myself - Let's calculate C(n,1), of course it is n . Put n=1/2 so C(1/2,1) is equal to 1/2 apply formula of combination C(1/2,1)= (1/2)!/{1!(-1/2)!} . Now knowing 1/2! as √π/2 , equate both equations and hence we get value of (-1/2)! as √π . Incredible . Similarly we can calculate some more negative and fractional factorials . If you know this trick already, then this trick has been already discovered, but if no one knows this trick then I am the first to use this .

[@RandomGuy-bf8wq]

The even more crazy thing about this problem is that a=2.33.... is the solution for making e^(x) - a tangent to ln x as well

[@peterojdemark]

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.

[@blackpenredpen]

Thanks! I actually recorded that video a few days ago. Here’s my Chinese version kzbin.info/www/bejne/fXW2Z4mNnMiIjtk and my English version will come out this week.

[@peterojdemark]

@@blackpenredpen Great! Looking forward too it:)

[@shivansh668]

Loved this innovative manipulation 🧡

[@mingmiao364]

Interesting problem: how many solutions does the equation a^x = log_a(x) have, where a>0? The answer, which depends on the range of a, is very intricate!

[@carultch]

Interesting question. To solve it, we'd need to set up a second equation to solve for the two unknowns. This would allow us to solve for the special value of a, where a^x = ln(x)/ln(a) has exactly 1 solution. Greater than this, there are no real solutions, and less than this, there are two solutions. If these two functions touch just once, they will also have the same derivative at the point where they touch, because they will both curve away from each other. This means we set their derivatives equal. d/dx a^x = ln(a)*a^x d/dx ln(x)/ln(a) = 1/(ln(a)*x) Our system of equations becomes: ln(a)*a^x = 1/(ln(a)*x) a^x = ln(x)/ln(a) Solve both equations to isolate a^x a^x = 1/(ln(a)^2 * x) a^x = ln(x)/ln(a) Equate them to each other: 1/(ln(a)^2 * x) = ln(x)/ln(a) Cancel one factor of ln(a): 1/(ln(a)*x) = ln(x) Multiply: 1 = x*ln(x)*ln(a) This equation will have a real solution, when ln(x) = 1, and x = 1/ln(a). This means that our solution is the following: x = e a = eth root of e, which is approximately 1.445.

[@carultch]

Continuing with this example, there will be two solutions, until a = 1, in which case you will have a degenerate case of a vertical line intersecting a horizontal line, at just one point. You wouldn't be able to solve for that one analytically, as you would get an error when you attempt to do so. The vertical line is x=1, and the horizontal line is x=1. For values of a that are less than 1, there is always just one real intersection of the two curves, when the logarithm curve is mirrored about the x-axis. That one solution will follow the diagonal line of y=x, where y=a^x and y=ln(x)/ln(a) are the two equations being graphed. This continues until another degenerate case at a=0, again where two perpendicular lines appear to meet at the origin. Although they don't really meet at the origin, because 0^0 is undefined.

[@anakin07]

I can’t believe I understood that. I’m not native English and didn’t have integrals in school yet. Your explanations are amazing❤️

[@IIBLANKII]

Not going to lie, I understood everything until you slapped W in there.

[@nasekiller]

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

[@axbs4863]

Finally someone decided to update math

[@mrborn1637]

i love that you're using version 5 of geogebra.

[@agr_]

You should do a part II problem where the function f(x) = ke^x touches ln(x)

[@theimmux3034]

Such a nice answer

[@catlilface]

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)

[@AlBoulley]

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.

[@narrawa2650]

Some interesting findings. The answer can be simplified to e^w(1)+w(1) as -ln(w(1) = w(1). Also this number, lets say a, can also be used to make them touch by shifting them vertically ([e^x] - a = lnx at a single point) as the function are mirrored over y=x. I actually solved the problem the vertical way first, then noticed my answer was the same as his, then found the reason why.

[@renatotafaj1507]

That was actually cool 😎

[@blackpenredpen]

Thanks!

[@nonono8108]

1:21 [keep in mind] only 1 *REAL* solution!

[@kanitatewari7604]

And if we take a>2.33 they intersect at two points

[@deedewald1707]

True to two points !

[@kutuboxbayzan5967]

i didn't watch your video, i think solution is a=w+1/w while w*e^(w)=1, its about to be 2.32 edit:Looks like my guess is true. The answer can simplify to w+1/w

[@khaledajlouni6419]

I want to ask you about solving the equation 1/x =ln(x) l took e for both sides and I divided by x taking the w Lambert function and take the reciprocal and I had a different answer why?

[@namanmishra703]

Yours is faster and gives the same answer. As he said in the video e^W(1)=1/W(1).

[@khaledajlouni6419]

@@namanmishra703 thanks I missed that

[@namanmishra703]

@@khaledajlouni6419 No problem. If you wonder why it is that way, it's really just the definition of the Lambert W function. x e^x = c implies x = W(c) Conversely, W(c) e^W(c) = c

[@Superman37891]

Or by inspection you could do a = e so that they meet at the point (e, 1)

[@laCOHSSA]

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

[@aniketeuler6443]

Very beautiful Steve sir keep uploading stuff like that Sir 😀

[@pojuantsalo3475]

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.

[@SIB1963]

Nice! And since e^[W(1)] = 1/W(1) - in other words, W(1) = e^-[W(1)] - then a = e^W(1) - ln[W(1)] = e^W(1) - ln[e^-[W(1)]] = e^W(1) - [-W(1)] = e^W(1) + W(1) Which also tells us that W(1) = -ln[W(1)]. Nicer!

[@Mothuzad]

Would you be interested in calculating the minimum distance between these curves? Similar in spirit to what you did here, but the steps should be completely different. Off the top of my head, you'd apply the Pythagorean theorem to both formulas, giving each an independent variable, then find the minimum of the resulting 2D function with calculus.

[@blackpenredpen]

Thanks for the comment. I forgot to put an old video on the minimum distance between e^x and ln(x) in the description. It is there now. 😃

[@Mothuzad]

Interesting! I hadn't noticed that nearest points on two curves would have to have equal derivatives, but it makes sense. Using the method I described, I think you'd end up finding a zero for the difference between the two derivatives, which ends up being the same fact.

[@edgardojaviercanu4740]

a beautiful exercise.

[@babajani3569]

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.

[@zildijannorbs5889]

e^(x^2+1) = pi^2x? Gosh, never let me become a math teacher, my exams would ruin lives

[@Shreyas_Jaiswal]

Finally revealed, BPRP is from U.S.

[@johannchevrier7063]

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 👏😉

[@hotlatte1222]

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～

[@blackpenredpen]

Hey Stanley, 這題別人也有問過, 我得好好想想!

[@saniya1180]

am I right ,Sir ln(X) +2and e^x also touch together and ln(X) +3 and e^x are intersect at two point.

[@64.maivananhtuan5]

Cảm ơn ad rất nhiều thank you very much i am from vietnam

[@shahadkftarawneh5431]

#الشيخ_جراح #انقذوا_حي_الشيخ_جراح #savesheikhjarrah #حي_الشيخ_جراح #لا_لتهويد_القدس #أنقذوا_حي_الشيخ_جراح #لن_نرحل

[@MathElite]

Amazing video

[@MessedUpSystem]

At this point I feel like this is Lambert W function the channel

[@kkkkjlkkkkj]

beautiful.

[@platformofscience9790]

Dear sir ,I have a question:Why do we use e for natural log or normal exponential functions.can you help me,please?

[@GammaStyleGaming]

It is named after Euler, the guy who discovered it. The number e is also called Euler's constant. I suggest checking out 3Blue1Brown, if you want to know how we get the value of e.

[@platformofscience9790]

I know that but I wanna know why do we use it.

[@Lucsji]

Normal exponential functions are of base e because it's the only base with which you can have exp'(x) = exp(x). This property of exp(x) is one of the things that make it so special, because it allowed the simplification of very complex math. As for logarithms, they were actually used by mathematicians before the exponetial functions were widely adopted. A Scottish mathematician named John Napier wanted to find a way to make hard computations easier by transforming products into sums. For that, he came up with the "logarithm tables", which you can read if you know the property log(a×b) = log(a) + log(b). In the table that John Napier came up with, he used the logarithm of base e, because it was the simplest form to do so. In a way, they made it base e without knowing.

[@Lucsji]

There are probably more details I ignored, but I hoped it helped.

[@platformofscience9790]

Surely it helped; Thank you for response like this,I appreciate it a lot,again thank you.

[@Nolord_]

a is also = to : 1/W(1) + W(1)

[@lukeroney7091]

I found the minimum value of ln(x)-e^x by setting the derivative equal to 0 and then plugged that x value in to ln(x)-e^x and got -2.33

[@Nidhi-ks6rn]

if it had a solution it would be like 2 parallel lines meeting together...ROFL

[@ronbannon]

I wrote a similar question for my students using 10^x = log (x), and used Newton's Method to solve for the horizontal shift. Video one is setting up the problem; video two is using Sage to solve the problem. 1. kzbin.info/www/bejne/oauokIKZqaqVhcUsi=qHLn8FRnUR0PfG64 2. kzbin.info/www/bejne/nIDKkpZud7B1nJosi=1HpNQSolnOflIjAy

[@justanalthere2187]

aye im happy i did it all mentally

[@madhavapai4321]

Why do e^(x-a) to shift e^x to the right? Why not (e^x)+a

[@yoav613]

Do you have any problem that you have no video related to it ?!

[@tambuwalmathsclass]

Really cool

[@willlenchus8166]

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)?

[@levskomorovsky1762]

e ^ x = x, The derivative of e^x is equal to the derivative of ln x e ^ x = 1/ x x e^x = 1 W (x e^x) = w ( 1) x = w (1) at the same time x = 1 / w (1) How is that?

[@tedchirvasiu]

When even a mathemarical formula finally meets someone but you don't

[@lucastellmarchi1948]

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.

[@asamenechbayissa553]

What if we move e^x to the right one unit and ln(x) up by one unit, e^(x - 1) is tangent to ln(x) + 1 and the post of intersection is (1 , 1)

[@klausolekristiansen2960]

e is e and ln is ln, and never the twain shall meet.