It's really annoying how correcting misinformation on a site like this always causes its Algorithm to promote the misinformation more than your correction
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.
@@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.
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.
@@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.
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
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_ )
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.
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).
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
> 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.
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 🙂
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 }]
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
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.
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
@@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.
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
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: \[ e^{W(\log(25))} \approx 3.2189 \]
Another mathematical solution video using (what I call the Willy Wanka function because of the plethora of TH-cam trick videos requiring its application) the Lambert W-function.
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
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
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.
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.
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 😂
@@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 :)
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.
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.
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.
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.
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.
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".
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
@@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”
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.
There is no Harvard entrance exam.
It's really annoying how correcting misinformation on a site like this always causes its Algorithm to promote the misinformation more than your correction
its called clickbait
I thought to get accepted to Harvard one has to simply express support for hamas or any other jihadist terror group. 😂😢
I bet the video creator pulled this from a qualifying exam and has no concept of what a QE even is...
There is no vega1447
I define the Skibbity Q function as returning x when applied to x^x. Therefore the answer is Q(25).
Lambert W is fr 🥲
@@brain_station_videos Lambert W and Skibbity Q are both equally real.
x^x has too many discontinuities and is unable to be broken into branches as the Lambert-W function is.
@robertanderson1043 I already made that function long ago and named it the mu function μ(x). So it's mine, not yours.
@@robertanderson1043Give me your rigorous definition and details of it. If not, get tf out.
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.
That's true. But atleast we are able to find a numerical value.
i think you can write your answer as an expression with a function
I'm guessing you hate logs and exponentials and regular sines/cosines/tangents and square roots and reciprocals too then?
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).
@@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.
1:20 sports
EA sports to the game
AHAHAHAHAHAHAHA
what kind of mind do you have. You *MADE ME HEAR THOSE WORDS WITH THE EA ANNOUNCER'S VOICE, DANG IT!!!* @.@
Nah, the guy def did it on purppse
i think it isnt log? the natural log(base e) is ln(x)
You are right. log is base 10 logarithm and ln is the natural logarithm(the base e logarithn)
I guess the solution is just to replace log be ln
It was a notation error. At least the narrator said "the natural log".
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.
@@t-cc3377jfnot error , difference .
It's still heavily used for natural log in many places .
3:43How do you calculate this?
using numerical methods
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.
W is a lesser known function but it is well-known enough for many mathematical softwares to have it as a built-in function
Newton's method
@@DrHyperionSundisliked for low self esteem
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
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
Can you please write ln(x) instead of log(x)? Natural log is not the same as log with base 10.
In American academia, log is implied as being base e unless otherwise specified
In the whole of calculus, we rarely use log with something else as base except for e
@@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.
This is the first time I have heard of the W Lambert Function. Though the procedure looks familiar.
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
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_ )
log is base 10, ln is base e
0 is a natural number in maths
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.
Ah and we find the computer scientist/applied maths person in the group
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.
Bruteforced in calculator 2,963219774895. It's pretty damn close :D
@@WilliamGerot Do I detect a hint of "us vs. them"?
@@pbierre Nah I just love the different approaches people take it adds positive diversity
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).
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
Adding to the reply by madden.
The lambert W function is defined. It was defined in this video and is taught at degree level.
> 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.
...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.
Are you in Harvard💀
youre saying it like any random can get into harvard
I mean if you are studying this and don't understand something then this video is a great tutorial
2:15 AAAAAAAH YES ! x is totally equal to e^log(x)
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 🙂
@az3224 yeah but no. He should have written e as the base of the log and not only "log(x)" otherwise is completely falsz
@@azizbronostiq2580 Some countries teach log instead of ln. (And it’s more satisfying to write)
@@partial-m4n yeah but it's still wrong
@@azizbronostiq2580 Not really wrong, advanced people use that
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 }]
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}
@@JUGNUMEHROTRANEETASPIRANToh genius over here
@@Nihalshanu22 Thanks 😁
Q: What's green and commutes
A: An abelian grape
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
-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
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
2.96324
See, the question was: x^x = 25. Then, can't we write ( 25 ) as :- 5^2. Then x=5 and upper x=2. Can we do this??
We can't since only a single value of x has to be on both sides
X has to be same number because there only one variable
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.
x^x = 25 = 5^2
xlnx = 2ln5
let y = lnx, e^y = x
ye^y = 2ln5
y = lnx = W(2ln5)
x = e^(W(2ln5))
I definitely need to use this equation while shopping at the grocery.
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
just seeing the problem makes me think about using this function
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
Ok, but where do I find the value of e^W(log(25))?
@@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.
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
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:
\[
e^{W(\log(25))} \approx 3.2189
\]
Just put this in a solver that uses bisection, run some iterations and done, easy. Solves this whole class of thumbnail problems.
But if you’re gonna use an external source like Wolfram anyways I’d just plot it in Desmos and intersect it
Now find every complex solution.
@@maddenbanh8033none
Those who use "log" as the base e logarithm, are following the contemporary trend and are in the cool club.
If you mean 'contemporary' as 1906. That construction is found on Boltzmann's tombstone.
In my country, it’s common to use log and we also learn like that
So, ye, I can agree with that
Another mathematical solution video using (what I call the Willy Wanka function because of the plethora of TH-cam trick videos requiring its application) the Lambert W-function.
I think you should explain the Lambert equation for this to be educational at all
You can use newton raphson method to avoid look up table.
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
Literally did the exact same thing....
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
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
Bit smaller than 3, that's what came first to my mind while looking at the equation.
2.963 gets really close.
2.9633 is approximately exact
@@itsmetanay"Approximately exact" is a funny combination of words.
@@itsmetanay2.96322 is closer
google says 2.963219774894 is close enough its a rounding error
@ Professor Google was always over the top! 🤣
Why do you keep calling log() with base 10 "a natural log" ?
Solution is only in case by using natural logarithm. Log is not natural
Why are there so many logs? Are we building a house?
Man Harvard really loves their Lambert W
I know, right? But it's really useful. I just wish it were easier to calculate its values.
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.
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.
Love the AI voice. "The Lambert double...(long pause)...u function..."
how could I live without W Lambert Function ?
cannot 💀
Amazing problem, scaring in the beginning but as you started explaining, the fog cleared and the brain sparkled.
I expected Lambert function to show up. I'm already used to seeing those videos XD
(x^x)' = x(x^(x-1)) = x^x which makes its taylor expansion very simple.
"And why do we need to know the answer to this?"
".. For...... uh... Science!"
Isn't Lambert W function only valid when w is a complex number?!
The solution is between 2 and 3. No need to be more precise on that 😅
Everyone in the comments are WRONG. The ACTUAL answer is: X = 2.96321977489346
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 😂
.....approximately..................................
@@chrupek439there's a decimal after 2 not a comma
@@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 :)
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.
So basically, you didn’t solve the problem. You just gave it a name.
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.
THIS IS 8TH GRADE EXPONENTS CHAPTER QUESTIONS IN INDIA , and theirs no harvard entrance exams
I actually found x 2.964 so im very proud of my self
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.
Me who thought it was 5^2 = 25 🥲
thats why i mentioned it in the video 🤣
That was my first thought then I attacked my calculator for the best approach: trial and error with guessing. 😎
Why people feel the need to solve using algebra over trial and error I’ll never understand.
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.
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)
If its not 5 I'm all out of ideas
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
For me , i never imagine a brut numer like 1 , i see wave and 1 in the top
Nice. It makes a lot for the planet
I figured it out in less time using a calculator and guessing and got more precision than the algebraic way
He used an arbitrary amount of precision.
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
Harvard entrance exam 🤡 JEE Advanced ☠️☠️
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.
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.
Or, you can solve this numerically, since you know the answer is between 2 and 3.
Have you heard the one about the mathematician and his logs. Well, he worked them out using a pencil.
This is essentially saying we define the solution x^x = y and F(y), wtf.
Now you have a new problem :
What is the value of W(ln(25)) ?
For which you will need a calculator 😂
So good....as always
Bruh just guess and check, i got x = 2.9634
x^x can be written as x X x and 25 can be written as 5 X 5 So x is 5
2.9634 approx
2^2=4 3^3=27 4^4=256 5^5=3125
and? what is purpose of this comment?
yea, just the click baiting shit as always
Absolute headache
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".
Calculators are not allowed - said all of my college math teachers.
I got 2.96322 in my head after a few minutes
2.963^2.963422111 is more accurate its about 24.9999999
i thought the answer for x to the power of x will have 0.50 as the value of x
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
No because log means log base e
@ that becomes ln
@@0943kt "log" means "log base e" in more conventional mathematics
@@adw1z i thot by default log means log base 10?
@@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”
Shouldn’t it be e^ln(x), instead of e^log(x)
I guess it is called lambert W function for a reason 💀
I will be a truck driver.
sitting and finding this thru trial and error also works.. if you do jack about W()
x = approx 2.96322
How is log(x) equal to ln(x)?
me be like -
1^1 = 1
2^2 = 4
3^3 = 27
so it would be approx ~2.9 something
OMG YES
same lol
I find that answer before the video ends,because i am an indian
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.
On peut utiliser Excel et la fonction : "valeur cible"
Дружище, ты не написал, как бы ты вычислил функцию Ламберта без всяких там wolfram и т.д. Это явно не примут на экзамене
Ok so..... i haven't learnt it yet but in advance what is log?
Forget it I gave up understanding it I will learn it in 11th or 12th then I will come back to this vid