Hardest Exponential Equation!

Hardest Exponential Equation! Math Olympiad
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[@brain_station_videos]

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[@Vega1447]

There is no Harvard entrance exam.

[@kabulykos]

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

[@thekaliss1]

its called clickbait

[@mightyoak11111]

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

[@thadwuj668]

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

[@thovenach]

There is no vega1447

[@robertanderson1043]

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

[@brain_station_videos]

Lambert W is fr 🥲

[@robertanderson1043]

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

[@WilliamGerot]

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

[@Nethuja_GunawardaneSL]

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

[@budderman3rd]

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

[@studentofspacetime]

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

[@Wutheheooooo]

Except you can calculate Lambert W function without any calculator now, it has a fomular, not closed form but you can approximate, it based on \ln x which can also approximate using newton's method. It is solved, you found a fomular to the solution.

[@Wynb824]

Not really, since for this kind of thing, you can't have a solution represented in elementary functions

[@williamstraub3844]

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]

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

[@tkyk314]

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

[@adw1z]

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

[@JeanG-s9j]

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]

​@@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]

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

[@3141minecraft]

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

[@shlok2444]

I guess the solution is just to replace log be ln

[@t-cc3377]

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

[@MD-kv9zo]

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]

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

[@StevenTorrey]

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

[@axumitedessalegn3549]

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]

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]

1:20 sports

[@math_solver_N]

EA sports to the game

[@tunistick8044]

AHAHAHAHAHAHAHA

[@7F0X7]

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

[@Krazykahaan]

Nah, the guy def did it on purppse

[@divyamkumar1339]

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

[@CAustin582]

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

[@death704]

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

[@edwolt]

​@@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]

​@@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

[@TechMoon-mz7fo]

​@@CAustin582No, it is not. I have three Algebra books and two Calculus books, and the Common Log 'log' is base 10, ALWAYS, and the natural log 'e' is either 'e' or lnx, ALWAYS Fk what your smoking

[@chouch]

...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]

Are you in Harvard💀

[@aurelianmasdrag2179]

youre saying it like any random can get into harvard

[@bakasteveuwu2822]

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

[@pizzaofdarkness4041]

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

[@pbierre]

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]

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

[@Brigadier_Beau]

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]

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

[@pbierre]

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

[@WilliamGerot]

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

[@andrewy664]

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]

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]

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]

> 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.

[@whiteeyesgamerright1954]

Let's Introduce jee advanced

[@PinoyRobots]

how could I live without W Lambert Function ?

[@brain_station_videos]

cannot 💀

[@donsimon2830]

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

[@KasyapH]

3:43How do you calculate this?

[@1-human]

using numerical methods

[@DrHyperionSun]

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]

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]

Newton's method

[@jaudatalhusen9049]

​@@DrHyperionSundisliked for low self esteem

[@abuelovinagres4411]

I am amazed by te amount of extra steps only to reach a result without any explanation.

[@BG-bq1qp]

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]

2.96324

[@thecompliationvideos4246]

-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

[@bakersbread104]

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

[@Wutheheooooo]

He did, it is the inverse function of xe^x

[@georgegrubbs2966]

Natural log is written as "ln". "log" implies "log base 10"

[@azizbronostiq2580]

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

[@az3224]

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]

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

[@laplace-fourier]

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

[@azizbronostiq2580]

@@laplace-fourier yeah but it's still wrong

[@laplace-fourier]

@@azizbronostiq2580 Not really wrong, advanced people use that

[@JUGNUMEHROTRANEETASPIRANT]

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]

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]

@@JUGNUMEHROTRANEETASPIRANToh genius over here

[@JUGNUMEHROTRANEETASPIRANT]

@@Nihalshanu22 Thanks 😁

[@Rone-q8v]

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

[@samus88]

The fuck is a Lambert W function?

[@just-dl]

2.963 gets really close.

[@itsmetanay]

2.9633 is approximately exact

[@netanelkomm5636]

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

[@labyrinth2646]

⁠@@itsmetanay2.96322 is closer

[@MUI_Noam12]

google says 2.963219774894 is close enough its a rounding error

[@just-dl]

@ Professor Google was always over the top! 🤣

[@Guidussify]

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

[@HottyHelen]

@@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]

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]

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: \[ e^{W(\log(25))} \approx 3.2189 \]

[@lool8421]

just seeing the problem makes me think about using this function

[@strangerfun7949]

Me :- hit and trial ( with common sense )

[@Rise6474]

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

[@chrupek439]

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]

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

[@oAnshul]

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

[@chrupek439]

@@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 :)

[@SnowyPlayer]

The answer is irrational. The exact value is e^(W(ln(25)))

[@cyruschang1904]

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

[@Zeddy27182]

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]

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]

log is base 10, ln is base e

[@aurelianmasdrag2179]

0 is a natural number in maths

[@adarsh-k5i]

@@aurelianmasdrag2179 really?

[@syamprasad4455]

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

[@Prime_Gamerzz]

double you 🗣🗣

[@proking1033s]

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

[@xjz9264]

between 2.9 and 3.0

[@funprop9004]

It also could just be √8.79

[@livewithals]

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

[@ilyashick3178]

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

[@Idkhowtogeometry]

I thought the number was 5 and the exponent is 2 lol😅

[@mohdabdulrahman2837]

2.96322

[@ivanmcauliff4597]

That would be the super-square root of 25, sometimes just called the "super root" of 25. No calculator that I know of has a super-root function - it's the one operation I know of that's actually easier to do on an ancient slide rule than it is to do on a calculator. We put the cursor hairline on 25 on the LL3 (e^x) scale. Then we move the slide bar left so that the right 1 is on that hairline. Finally, we move the cursor hairline to the left and try to find the point on the LL3 (e^x) scale where it matches with the point on the CI (1/x) scale. And that's roughly 2.96.

[@thorliebhammer7238]

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

[@robertwarren4734]

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

[@laplace-fourier]

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

[@foodymshmshinfomshart4000]

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

[@strengthinnumberstutoring61]

To get into Har-VARD, I simply apply the “Become a Harvard Stu-…DENT” function! This takes input of “me” and returns “Har-VARD Stu-…DENT!” Quod erat faciendum.

[@JonJenkins1982]

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

[@maddenbanh8033]

He used an arbitrary amount of precision.

[@wren51615]

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

[@elyesdimassi36]

approximately 2.963

[@rangorhodeo]

I would just take the calculator and do hit and trail of self multiplication

[@RunItsTheCat]

Man Harvard really loves their Lambert W

[@sanchellewellyn3478]

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

[@nagendrababugalibu7322]

I know logarithms perfectly

[@aksiiska9470]

elon musk would say "3power3=27" and i am almost there

[@highlyeducatedtrucker]

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

[@moulibratasarkar841]

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

[@odysseus9941]

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.

[@yuki7951]

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

[@louiscarl7629]

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

[@Dr_piFrog]

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.

[@hardik-venturepiano1962]

Lambert W function makes no sense to me. After all you HAVE to use a calculator to get your answer, Then why not just type in the equation and get your answer

[@harikeshavraman5506]

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]

Literally did the exact same thing....

[@odysseus9941]

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]

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.

[@Souf.df.90]

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

[@ЭльЯвор]

3**3 = 27 It is all you need

[@netravelplus]

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

[@hereticalgames3695]

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

[@hereticalgames3695]

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.

[@alexzuma2025]

the natural logarithm is written using ln. not log.

[@ryanhollstein4164]

Super square root of 25 which is a transcendental number

[@niarai6898]

Isn't it 5?

[@maroly8342]

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

[@laplacia]

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

[@Antonio-v2j]

Me who thought it was 5^2 = 25 🥲

[@brain_station_videos]

thats why i mentioned it in the video 🤣

[@just-dl]

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

[@mathguy37]

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

[@jim2376]

x = e^W(ln 25) Then Wolfram Alpha x = 2.96321

[@noobspidey8750]

You could also do it with linear approximation right?

[@viveksmenon123]

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

[@Slash1066]

If its not 5 I'm all out of ideas

[@saminyead1233]

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

[@Dev-b4d6x]

x is simply 5

[@randerson4009]

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.

[@johnjr2jr2]

Nice. It makes a lot for the planet

[@SuryaKant-u4h]

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

[@fomrehkuda]

x^x cant be written like x • x

[@shahabuddinansari9235]

Answer is approximately or may tell Answer is absolutely wrong

[@Grassmpl]

Definitely get out your lambert W notation

[@donsimon2830]

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

[@carloalbertocolaiacovo4182]

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

[@b213videoz]

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

[@sohayb1m-582]

I actually found x 2.964 so im very proud of my self

[@Boyscrazy719]

Harvard entrance exam 🤡 JEE Advanced ☠️☠️

[@DailyWorkoutEnjoyer]

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

[@DeepakKumar-fi4gp]

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

[@АлександрБедин-х2ш]

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.

[@RyanLewis-Johnson-wq6xs]

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

[@Why553-k5b_1]

and? what is purpose of this comment?

[@ronaldbryant5215]

Calculators are not allowed - said all of my college math teachers.

[@pingkai]

This is essentially saying we define the solution x^x = y and F(y), wtf.

[@kateknowles8055]

x = approx 2.96322

[@eclxpse2]

Bruh just guess and check, i got x = 2.9634

[@Esterified80]

Absolute headache

[@vatsalmakol7]

2.9634 approx