Hardest Exponential Equation!

Рет қаралды 431,663

Brain Station

Күн бұрын

Пікірлер: 448

@Vega1447 Ай бұрын

There is no Harvard entrance exam.

@kabulykos Ай бұрын

It's really annoying how correcting misinformation on a site like this always causes its Algorithm to promote the misinformation more than your correction

@kaliss6110 Ай бұрын

its called clickbait

@mightyoak11111 Ай бұрын

I thought to get accepted to Harvard one has to simply express support for hamas or any other jihadist terror group. 😂😢

@thadwuj668 Ай бұрын

I bet the video creator pulled this from a qualifying exam and has no concept of what a QE even is...

@thovenach Ай бұрын

There is no vega1447

@robertanderson1043 Ай бұрын

I define the Skibbity Q function as returning x when applied to x^x. Therefore the answer is Q(25).

@brain_station_videos Ай бұрын

Lambert W is fr 🥲

@robertanderson1043 Ай бұрын

@@brain_station_videos Lambert W and Skibbity Q are both equally real.

@WilliamGerot Ай бұрын

x^x has too many discontinuities and is unable to be broken into branches as the Lambert-W function is.

@Nethuja_GunawardaneSL Ай бұрын

@robertanderson1043 I already made that function long ago and named it the mu function μ(x). So it's mine, not yours.

@budderman3rd Ай бұрын

@@robertanderson1043Give me your rigorous definition and details of it. If not, get tf out.

@williamstraub3844 Ай бұрын

This is why I hate the Lambert W function! It's like saying "What is the solution to x = sin(37)? Why, it's arcsin(x) = 37." The W function is useless unless you have a calculator or Wolfram Alpha handy.

@brain_station_videos Ай бұрын

That's true. But atleast we are able to find a numerical value.

@0943kt Ай бұрын

i think you can write your answer as an expression with a function

@adw1z Ай бұрын

I'm guessing you hate logs and exponentials and regular sines/cosines/tangents and square roots and reciprocals too then?

@JeanG-s9j Ай бұрын

I agree, its like if I could create my own function, lets say J(x^x)=x, then if x^x=25, the solution is x=J(25).

@TheMathManProfundities Ай бұрын

@@JeanG-s9jBut you wouldn't be able to get a value from that. It's more like x=√2. It had a value but you basically need computation to evaluate it to any significant level.

@SVMNSP6213YT 2 ай бұрын

i think it isnt log? the natural log(base e) is ln(x)

@3141minecraft 2 ай бұрын

You are right. log is base 10 logarithm and ln is the natural logarithm(the base e logarithn)

@shlok2444 2 ай бұрын

I guess the solution is just to replace log be ln

@t-cc3377 2 ай бұрын

It was a notation error. At least the narrator said "the natural log".

@MD-kv9zo Ай бұрын

Apparently a lot of people(including my maths and physics teachers)do that even though it just gets more confusing. Log should be to the base 10 and ln to the base e.

@FundamSrijan Ай бұрын

@@t-cc3377jfnot error , difference . It's still heavily used for natural log in many places .

@divyamkumar1339 Ай бұрын

Can you please write ln(x) instead of log(x)? Natural log is not the same as log with base 10.

@CAustin582 Ай бұрын

In American academia, log is implied as being base e unless otherwise specified

@death704 Ай бұрын

In the whole of calculus, we rarely use log with something else as base except for e

@edwolt 29 күн бұрын

@@death704In my calculus course, when talking about natural log we used ln. But for some reason, in CS courses people used just log as meaning log2 when they could've be using lb, which is kinda confusing.

@Mk3737 6 күн бұрын

@@edwolt yeahh I'm a high school student and generally we're aware of ln and log still we use log instead of ln cuz yeah we've to rarely use log with base 10 unless until specified in the question

@axumitedessalegn3549 Ай бұрын

A better answer is 2.9632 and you can just do log(25)/log(x) in a graphing calculator and look for a value that is x=y

@tsolanoff Ай бұрын

It’s a bit confusing to require applicants to know advanced math (Lambert W function)which is supposed to be taught in higher education institutions

@user-ix9zu5es6j Ай бұрын

1:20 sports

@math_solver_N Ай бұрын

EA sports to the game

@tunistick8044 Ай бұрын

AHAHAHAHAHAHAHA

@7F0X7 Ай бұрын

what kind of mind do you have. You *MADE ME HEAR THOSE WORDS WITH THE EA ANNOUNCER'S VOICE, DANG IT!!!* @.@

@Krazykahaan 28 күн бұрын

Nah, the guy def did it on purppse

@StevenTorrey 2 ай бұрын

This is the first time I have heard of the W Lambert Function. Though the procedure looks familiar.

@pbierre Ай бұрын

It's ~ 2.96322. There is a more modern, easier to understand approach based on successive approximation and iterative computation. All you need is a calculator with exponentiation function x^y.

@WilliamGerot Ай бұрын

Ah and we find the computer scientist/applied maths person in the group

@Brigadier_Beau Ай бұрын

I used estimation and managed to get that it would be close to but less than 2.965. I didn't feel like refining further.

@gergokarath5739 15 күн бұрын

Bruteforced in calculator 2,963219774895. It's pretty damn close :D

@pbierre 15 күн бұрын

@@WilliamGerot Do I detect a hint of "us vs. them"?

@WilliamGerot 14 күн бұрын

@@pbierre Nah I just love the different approaches people take it adds positive diversity

@bakersbread104 29 күн бұрын

I think you should explain the Lambert equation for this to be educational at all

@andrewy664 Ай бұрын

Hello, 2 errors here: (1) you cannot just apply an arbitrary (in this case -- even undefined) function to both parts of an equation without proving than you don't lose any root and don't introduce any new root; (2) the same, when author applies exponential function to both sides of the equation (particularly, x=0 becomes a part of the domain after this operation).

@maddenbanh8033 Ай бұрын

I don't see the problem here, if he's only focusing on the reals then clearly he's only using the principal branch and secondly while you werent exactly clear, 0 was defined at the start

@Munnstery Ай бұрын

Adding to the reply by madden. The lambert W function is defined. It was defined in this video and is taught at degree level.

@andrewy664 Ай бұрын

> clearly he's only using the principal branch @maddenbanh8033 , it is not a problem in this particular equation, but it may be a problem in another expression. I consider this video it to be an educational one, so as a newbie I'd like to to see a precise and systematic approach here, not just a green way. Moreover, a student may get a penalty for not mentioning those facts and possibly introducing inequivalence. And all my initial points relate to the real numbers set only. @benmunn7481, the function is of course defined by itself, but not in the scope of this task. I've studied dozens of higher math disciplines for over 5 years, but have never heard aout it, so it's hard to say it is widely used here and there. My nitpick was that for some functions it is OK to do so, but for another we may lose equivalence during the solution. E.g. if we use f(t) = 0 instead of Lambert W function, we will end up with the solution that x may be any real number which is, obviously, an error.

@azizbronostiq2580 Ай бұрын

2:15 AAAAAAAH YES ! x is totally equal to e^log(x)

@az3224 Ай бұрын

Log(x) to the base e tells us what should be the power of e to get x For example log2 to the base e equals 0.30103 this means e raised to power 0.30103 is equal to 2 Similarly log x to the base e is the power to which e is raised to get x Therefore x =e^logx Hope it helps you 🙂

@azizbronostiq2580 Ай бұрын

@az3224 yeah but no. He should have written e as the base of the log and not only "log(x)" otherwise is completely falsz

@prof._andrew_wiles Ай бұрын

@@azizbronostiq2580 Some countries teach log instead of ln. (And it’s more satisfying to write)

@azizbronostiq2580 Ай бұрын

@@prof._andrew_wiles yeah but it's still wrong

@prof._andrew_wiles Ай бұрын

@@azizbronostiq2580 Not really wrong, advanced people use that

@KasyapH Ай бұрын

3:43How do you calculate this?

@1-human Ай бұрын

using numerical methods

@DrHyperionSun Ай бұрын

I would try Newton's method assuming z=W(ln(25)) and Derivating f(z). But this is because I am dumb, sure there are more pretty methods.

@latorredelreloj Ай бұрын

W is a lesser known function but it is well-known enough for many mathematical softwares to have it as a built-in function

@tunistick8044 Ай бұрын

Newton's method

@jaudatalhusen9049 25 күн бұрын

@@DrHyperionSundisliked for low self esteem

@Zeddy27182 Ай бұрын

The log(x) is not necessarily base 10. It depends on how it is defined! High school: base 10 College Math: base e Python, R, etc : base e CS : base 2 Even the definition of natural number varies: Math: 1, 2, 3, ... CS : 0, 1, 2, ... Overall, it really doesn't matter at all. "The essence of mathematics lies in its freedom." - Cantor

@cdmcfall Ай бұрын

ISO 80000-2, which is supposed to be international notation standards, says this: logₐ _x_ => logarithm to the base _a_ of _x_ ; standard, unambiguous notation ln _x_ = logₑ _x_ lg _x_ = log₁₀ _x_ => This was formally just log _x_ lb _x_ = log₂ _x_ log _x_ => This should only be used when the base does not need to be specified (most calculators treat it as log₁₀ _x_ while many apps, including Wolfram Alpha, treat it as ln _x_ )

@alexandreclergeaud4672 28 күн бұрын

log is base 10, ln is base e

@aurelianmasdrag2179 28 күн бұрын

0 is a natural number in maths

@adarsh-k5i 10 сағат бұрын

@@aurelianmasdrag2179 really?

@pizzaofdarkness4041 Ай бұрын

I definitely need to use this equation while shopping at the grocery.

@GwnTim1 Ай бұрын

But if you’re gonna use an external source like Wolfram anyways I’d just plot it in Desmos and intersect it

@maddenbanh8033 Ай бұрын

Now find every complex solution.

@CMANIZABALLER Ай бұрын

@@maddenbanh8033none

@BG-bq1qp Ай бұрын

2.9632 just do 2.5^2.5 and go up until you are above 25, then give it more decimals until you’re close enough

@wbytgaming2843 26 күн бұрын

2.96324

@donsimon2830 Ай бұрын

Q: What's green and commutes A: An abelian grape

@moulibratasarkar841 4 күн бұрын

If you really just want approximate values, there are simpler solutions not using Lambert s w function

@PinoyRobots Ай бұрын

how could I live without W Lambert Function ?

@brain_station_videos Ай бұрын

cannot 💀

@Dr_piFrog 18 күн бұрын

Another mathematical solution video using (what I call the Willy Wanka function because of the plethora of KZbin trick videos requiring its application) the Lambert W-function.

@lool8421 Ай бұрын

just seeing the problem makes me think about using this function

@syamprasad4455 18 күн бұрын

You can use newton raphson method to avoid look up table.

@JUGNUMEHROTRANEETASPIRANT Ай бұрын

I solved it in my mind as follows : X ln{x}=ln[25] Or ==> e^ln{x} . ln{x} = 2ln[5] Therefore, x =e^W{2ln[5]}. [OBVIOUS WHY OR EXPLAINATION GIVEN IN THE VEDIO{about W function }]

@JUGNUMEHROTRANEETASPIRANT Ай бұрын

If one knows about W , it shall take him/her less than 1 min to solve it mentally , otherwise , one would make guesses{e.g here , x~3 is a guess}

@Nihalshanu22 Ай бұрын

@@JUGNUMEHROTRANEETASPIRANToh genius over here

@JUGNUMEHROTRANEETASPIRANT Ай бұрын

@@Nihalshanu22 Thanks 😁

@alexzuma2025 10 күн бұрын

the natural logarithm is written using ln. not log.

@mathguy37 28 күн бұрын

There's a different interesting way i found that appears to approximate a value without using the lambert W function so take the log of 25, the base matters to keep the number real, so just make a reasonable guess. Then just take the log of 25 with that as the base. Keep recursively doing that and the answer will approach the solution It takes a lot of logs to converge though, so it's easier to use recursive functions if inputting this into a calculator. (or using ans) 100 logarithms gives ~2.96321977726 with a starting base of 3, which is very close to the true answer. Any starting base within a reasonable range gives a nearly equivalent answer as the number of logarithms increase. Using 5 gives 2.96321987478. I couldn't figure out range of bases that work though

@viveksmenon123 Ай бұрын

i have no idea about lamberts function. If I just want to approximate, I can just do a binary search between 2 and 3. 2.5^2.5, 2.75^2.75, 2.825^2.825, 2.93^2.93, 2.96^2.96

@odysseus9941 28 күн бұрын

Die Gleichung x^x=25 kann nicht mit einfachen algebraischen Methoden gelöst werden, da x sowohl als Basis als auch als Exponent auftritt. Stattdessen wird sie mithilfe von numerischen Methoden oder der Lambert-W-Funktion gelöst.

@DeepakKumar-fi4gp 26 күн бұрын

To solve the equation , let us proceed step-by-step: Step 1: Rewrite the equation We have: x^x = 25 \ln(x^x) = \ln(25) Using the logarithmic property , this becomes: x \ln(x) = \ln(25) Step 2: Approximation or numerical method The equation does not have a closed-form solution and must be solved numerically. Let's proceed: , so the equation becomes: x \ln(x) = 3.2189 Step 3: Estimate the solution Try values for : If , (too large). If , (close to 3.2189). The solution is slightly above . Step 4: Refine using numerical methods Using numerical tools (like Newton's method), we find: x \approx 2.559 Final Answer: x \approx 2.559

@samus88 Ай бұрын

The fuck is a Lambert W function?

@studentofspacetime 24 күн бұрын

So basically, you didn’t solve the problem. You just gave it a name.

@thorliebhammer7238 Ай бұрын

Those who use "log" as the base e logarithm, are following the contemporary trend and are in the cool club.

@robertwarren4734 Ай бұрын

If you mean 'contemporary' as 1906. That construction is found on Boltzmann's tombstone.

@prof._andrew_wiles Ай бұрын

In my country, it’s common to use log and we also learn like that So, ye, I can agree with that

@Guidussify Ай бұрын

Ok, but where do I find the value of e^W(log(25))?

@HottyHelen Ай бұрын

@@Guidussify I wonder the same thing, as far as I can see this isn’t a function like sin or log, there’s no however long formula to find W of a value. I doubt it’s a button on any calculator you can buy. So you need to either use an analytical approximation, numerical methods like Newton Ralphson or software that probably use that. I just used Excel’s goal seek facility.

@cdmcfall Ай бұрын

Iterative methods (brute force), mostly, or just run it through a Lambert W calculator. To evaluate this one, you would type in " e^(W_0(log(25))) " into the Wolfram Alpha search bar. Make sure you take a look at the graphs of y = x^x and y = 25. These sometimes have real solutions that aren't immediately obvious, as in the case of 2^x = x^2

@lanaforeal2588 Ай бұрын

To find the value of \( e^{W(\log(25))} \), we can utilize the property of the Lambert W function, which states that if \( y = W(x) \), then \( x = y e^y \). 1. First, compute \( \log(25) \): \[ \log(25) = \log(5^2) = 2 \log(5) \] 2. Next, we find \( W(\log(25)) \). Since \( W(x) \) is the function that satisfies \( x = W(x)e^{W(x)} \), we need to express \( \log(25) \) in a suitable form for the Lambert W function. 3. However, we can also use the property: \[ e^{W(x)} = \frac{x}{W(x)} \] For our case, this means: \[ e^{W(\log(25))} = \frac{\log(25)}{W(\log(25))} \] 4. Since \( e^{W(x)} \) simplifies to \( x \) if \( x \) is of the form \( y e^y \), we conclude that: \[ e^{W(\log(25))} = \log(25) \] Thus, the value of \( e^{W(\log(25))} = \log(25) \). If you need a numerical approximation, it can be calculated as follows: \[ \log(25) \approx 3.2189 \] So the final result is: \[ e^{W(\log(25))} \approx 3.2189 \]

@whiteeyesgamerright1954 7 күн бұрын

Let's Introduce jee advanced

@randerson4009 Ай бұрын

Why not just use an iterative approximation method on the original equation to the precision desired? This avoids the rearranging of the equation and finding the value of the resulting Lambert W.

@foodymshmshinfomshart4000 10 күн бұрын

Answer will be 2.9633 by hit and trial method using calculator. Estimated time 3 minutes..

@ilyashick3178 Ай бұрын

Solution is only in case by using natural logarithm. Log is not natural

@just-dl Ай бұрын

2.963 gets really close.

@itsmetanay Ай бұрын

2.9633 is approximately exact

@netanelkomm5636 Ай бұрын

@@itsmetanay"Approximately exact" is a funny combination of words.

@labyrinth2646 Ай бұрын

⁠@@itsmetanay2.96322 is closer

@MUI_Noam12 Ай бұрын

google says 2.963219774894 is close enough its a rounding error

@just-dl Ай бұрын

@ Professor Google was always over the top! 🤣

@odysseus9941 28 күн бұрын

Die Harvard University hat keine spezifische Aufnahmeprüfung wie z. B. eine standardisierte Prüfung, die alle Bewerber bestehen müssen. Stattdessen basiert das Aufnahmeverfahren auf einer ganzheitlichen Bewertung der Bewerbungsunterlagen, wobei viele Faktoren berücksichtigt werden.

@odysseus9941 28 күн бұрын

@yuki7951 Ай бұрын

I expected Lambert function to show up. I'm already used to seeing those videos XD

@RunItsTheCat Ай бұрын

Man Harvard really loves their Lambert W

@sanchellewellyn3478 Ай бұрын

I know, right? But it's really useful. I just wish it were easier to calculate its values.

@strangerfun7949 6 күн бұрын

Me :- hit and trial ( with common sense )

@Rise6474 Ай бұрын

Everyone in the comments are WRONG. The ACTUAL answer is: X = 2.96321977489346

@chrupek439 29 күн бұрын

2.963219774893456328309^2.963219774893456328309 = 25.000000000000000000159053596577 so nope, still working on it i got to: 2.9632197748934563283059504789757^2.9632197748934563283059504789757 = 25.000000000000000000000000000001 my calc wont let me add more numbers 😂

@kateknowles8055 27 күн бұрын

.....approximately..................................

@oAnshul 27 күн бұрын

@@chrupek439there's a decimal after 2 not a comma

@chrupek439 26 күн бұрын

@@oAnshul in polish schools we are taught to use comma, for larger numbers we just leave space between every last three digits. But I changed it for you anyway :)

@thecompliationvideos4246 18 күн бұрын

-Newton-Raphson method -formula: xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ) -Initial guess: x₀ = 3 1. f(3) = 3ˣ - 25 ≈ 2 f'(3) = 3ˣ (ln(3) + 1) ≈ 56.66 x₁ = 3 - (2 / 56.66) ≈ 2.9647 2. f(2.9647) = 2.9647ˣ - 25 ≈ 0.0775 f'(2.9647) = 2.9647ˣ (ln(2.9647) + 1) ≈ 52.33 x₂ = 2.9647 - (0.0775 / 52.33) ≈ 2.96322 (practically solved already, but I continued until it converged for 6 decimal places, tedious but simple) 3. f(2.96322) = 2.96322ˣ - 25 ≈ 0.00013 f'(2.96322) = 2.96322ˣ (ln(2.96322) + 1) ≈ 52.16 x₃ = 2.96322 - (0.00013 / 52.16) ≈ 2.9632198 4. f(2.9632198) = 2.9632198ˣ - 25 ≈ 0 x₄ ≈ 2.9632198

@carloalbertocolaiacovo4182 Ай бұрын

Bro you forget the modul inside the natural log when you do ln x^x = x ln |x| So you resolve that for x>o and for x

@livewithals Ай бұрын

Bit smaller than 3, that's what came first to my mind while looking at the equation.

@chouch Ай бұрын

...or why you should not study maths at harvard, nor watch random videos that pretend to be of mathematical nature. What a waste of time.

@SachinGupta-h7y Ай бұрын

Are you in Harvard💀

@aurelianmasdrag2179 28 күн бұрын

youre saying it like any random can get into harvard

@bakasteveuwu2822 21 күн бұрын

I mean if you are studying this and don't understand something then this video is a great tutorial

@harikeshavraman5506 Ай бұрын

Obviously in modern world, we don't need this method to solve equations in daily basis - Excel Goal seek is going to help you out solve this one - Or if it's necessary to do it by a normal calculator, We know that the number and the power variable should be the same... 1^1=1, 2^2=4, 3^3=27.....so x is somewhere near to 3 in order to get x^x=25... Do some trial & error, Try with x= 2.9 & 2.95,gives value as 21.9 & 24.3...so raise x to 2.96, gets 24.84 which is closer... Try 2.965,gets 25.1....try 2.963,gets 24.98....final try with 4 decimals....2.9633,gets 25......Believe me guys this just took me 2mins to do! I work on these equations on a daily basis in my work and I always prefer doing trial & error methods.... For complex eqns, we can use some numerical int methods like Simpson rule, etc to calculate x sooner

@mohitp66448 Ай бұрын

Literally did the exact same thing....

@b213videoz Ай бұрын

Why do you keep calling log() with base 10 "a natural log" ?

@netravelplus Ай бұрын

Amazing problem, scaring in the beginning but as you started explaining, the fog cleared and the brain sparkled.

@proking1033s 22 күн бұрын

Its simple 2 power 2 is 4 3 power 3 is 27 Answer should be closer to 2.8 or 2.9 Thats how objective questions work

@highlyeducatedtrucker Ай бұрын

Love the AI voice. "The Lambert double...(long pause)...u function..."

@Souf.df.90 27 күн бұрын

For me , i never imagine a brut numer like 1 , i see wave and 1 in the top

@AndrewUnruh Ай бұрын

OK...I guess my problem with the solution is this...There is no closed form solution of the Lambert W function so we have to use numerical methods. But if that is the case, I can just use a numerical method to solve for the original equation without the use of the Lambert W function, right? I guess the advantage is that if you don't know how to write a numerical solution, you can use an on-line Lambert W function calculator.

@Rudrakunjir 26 күн бұрын

Shouldn’t you be using ln for natural log. Log is for base 10. Atleast I think. Cus I got confused at the lambert w part.

@laplacia Ай бұрын

(x^x)' = x(x^(x-1)) = x^x which makes its taylor expansion very simple.

@0943kt Ай бұрын

2:16 im pretty sure x is not equal to e^log(x). if we apply ln, that gives us ln(x)=log(x), which the only solution to this is x=1. you need to use ln instead of log at the start so e and ln cancels

@adw1z Ай бұрын

No because log means log base e

@0943kt Ай бұрын

@ that becomes ln

@adw1z Ай бұрын

@@0943kt "log" means "log base e" in more conventional mathematics

@0943kt Ай бұрын

@@adw1z i thot by default log means log base 10?

@adw1z Ай бұрын

@@0943kt well it depends on context and what the usage of it is. Majority of the time in published papers, log will mean natural log. I found it’s mainly in school that they distinguish between ln and log. The reason is log base e is used pretty much everywhere all the time, whereas logs in other bases are very rarely used. It can be confusing, for example one of my courses used log, which actually meant log base 2 implicitly. So just be wary of the context of the problem; here it’s pretty obvious log meant base e, and he specified by saying “natural log”

@cyruschang1904 Ай бұрын

x^x = 25 = 5^2 xlnx = 2ln5 let y = lnx, e^y = x ye^y = 2ln5 y = lnx = W(2ln5) x = e^(W(2ln5))

@Rone-q8v Ай бұрын

Why are there so many logs? Are we building a house?

@wren51615 Ай бұрын

I saw the thumbnail and knew it was a little under three, and thought “oh god is it eulers number again”. Glad it wasn’t

@DailyWorkoutEnjoyer Ай бұрын

"And why do we need to know the answer to this?" ".. For...... uh... Science!"

@louiscarl7629 Ай бұрын

Just put this in a solver that uses bisection, run some iterations and done, easy. Solves this whole class of thumbnail problems.

@johnjr2jr2 Ай бұрын

Nice. It makes a lot for the planet

@JonJenkins1982 Ай бұрын

I figured it out in less time using a calculator and guessing and got more precision than the algebraic way

@maddenbanh8033 Ай бұрын

He used an arbitrary amount of precision.

@sheltondany8209 27 күн бұрын

sitting and finding this thru trial and error also works.. if you do jack about W()

@luclacourse424 Ай бұрын

the only thing that doesnt make sense to me in this equation is the use of x for 2 var...in programmation language x = 5 exponential x = var

@SuryaKant-u4h 12 күн бұрын

x^x can be written as x X x and 25 can be written as 5 X 5 So x is 5

@bowiebrewster6266 Ай бұрын

Lambert w function is cheating. Might aswell say i have the bowie-w function is the solution to a = x^x. So we get bowie(25)

@hereticalgames3695 28 күн бұрын

Why people feel the need to solve using algebra over trial and error I’ll never understand.

@hereticalgames3695 28 күн бұрын

Edit: 5^2 =25 so the range must be 2-5 3^3=27 so the range is 2-3 punch in like 2.9 and you’ll find it short 2.96-2.97 is the next range how many decimals do you practically need. It turns into simple busy work fast.

@АлександрБедин-х2ш Ай бұрын

What is the hardness of the task? If it is known in advance in which functions it is allowed to give an answer, then it is solved in 3 lines for any 9th graders.

@Mr.Icecream Ай бұрын

THIS IS 8TH GRADE EXPONENTS CHAPTER QUESTIONS IN INDIA , and theirs no harvard entrance exams

@DexlabHurts 21 күн бұрын

Isn't Lambert W function only valid when w is a complex number?!

@psydhant Ай бұрын

How is log(x) equal to ln(x)?

@maroly8342 28 күн бұрын

The solution is between 2 and 3. No need to be more precise on that 😅

@donsimon2830 Ай бұрын

Have you heard the one about the mathematician and his logs. Well, he worked them out using a pencil.

@7F0X7 Ай бұрын

Thanks for the video. But I took calc 1 in highschool and the "W" function was never taught. I highly doubt such a niche constant function would be part of any entrance exam except maybe those chinese entrance exams they give to poor, rural students for the express purposes of lowering their pass rates so the rich urban kids can dominate (this actually happens, look it up). You were too fast & loose as you alternated between saying "log" and "natural log" yet only writing "log". The entire problem should have been "natural log" only and written as "ln".

@l14debneelbarik75 8 күн бұрын

2.9634 is the answer. Done by hit and trial method 🎉

@Antonio-v2j 2 ай бұрын

Me who thought it was 5^2 = 25 🥲

@brain_station_videos 2 ай бұрын

thats why i mentioned it in the video 🤣

@just-dl Ай бұрын

That was my first thought then I attacked my calculator for the best approach: trial and error with guessing. 😎

@Esterified80 Ай бұрын

Absolute headache

@Realalexandro Ай бұрын

f(x)=x^x IS NOT a purely INCREASING function because on the half-interval of (0;1/e] it effectively decreases. You can calculate its derivative to confirm that. Next given that lim f(x->0) = 1 (which can be proved using L'Hospital's rule, also known as Bernoulli's rule that allows evaluating limits of indeterminate forms using derivatives), you should at least say that the potential max(f(x)) on (0;1/e] is 1, although it can't be achieved because 0^0 is an undefined expression! So on the decreasing "end" f(x) < 25 and can not have any solutions. On (1/e; +infinity) though f(x) is monotonously increasing, so on this place of the plot there can be no more than one solution that you've actually found. IMHO Harvard guys are pretty dumb to ask such questions, cause transcendent equations of such nature in general form CAN BE SOLVED ONLY USING W-Lambert function (which can't be represented in elementary functions) and moreover if you know this W-Lambert technique once and for all times all of these tasks are usually pretty easy to solve in terms of W-Lmb function. So producing correct solution only demonstrates that you know what W-Lambert is and how to apply it directly, nothing more! No guess, no creativity or originality of thought process here needed.

@RyanLewis-Johnson-wq6xs Ай бұрын

2^2=4 3^3=27 4^4=256 5^5=3125

@Why553-k5b_1 Ай бұрын

and? what is purpose of this comment?

@vatsalmakol7 Ай бұрын

2.9634 approx

@aoichan4353 Ай бұрын

i thought the answer for x to the power of x will have 0.50 as the value of x

@chair1694 Ай бұрын

So good....as always

@99Unknown Ай бұрын

Can you solve IIT physics questions ☠️☠️

@saminyead1233 Ай бұрын

Or, you can solve this numerically, since you know the answer is between 2 and 3.

@BiagioLatufara Ай бұрын

Shouldn’t it be e^ln(x), instead of e^log(x)

@thedungeonmaster8846 Ай бұрын

I thought it would be 5 ^ 2

@Slash1066 Ай бұрын

If its not 5 I'm all out of ideas

@FallenImmortal69 Ай бұрын

Ok so..... i haven't learnt it yet but in advance what is log?

@FallenImmortal69 Ай бұрын

Forget it I gave up understanding it I will learn it in 11th or 12th then I will come back to this vid

@fuxtnegrx Ай бұрын

You didn't explain lambert w function properly.

@brain_station_videos Ай бұрын

Yeah I have shown its use case (what it does). It was just a quick overview to solve this question! Probably in some future video I will talk about lambert W function in detail

@fuxtnegrx Ай бұрын

@brain_station_videos its not what I read in wiki. The fornula as you show it doesn't make sense 🤷🏻‍♂️

@napoleonbonaparte5783 Ай бұрын

@@fuxtnegrx He actually is correct as far as I know and the formula is correct but obviously has more complex applications in more advanced problems. And by the way if you have an issue with something he explained don't just say just say your wrong or you don't make sense. Bring forth what you understand and challenge what he said so that he can explain further or you lot can debate about it. That's more constructive than what you are doing right now. Anyway that's my two cents✌🏽

@Vega1447 Ай бұрын

Just solve x * log(x)= log 25. Using Newtons method x=x-(x*log(x)-l25)/(1+log(x)) starting with x=3 you get 10 digit accuracy after 2 or 3 iters. x=2.963219774893456.