Really easy problem. 90% of those 96% probably weren't able to solve, because they don't know W Lambert function. Thanks for bringing it for all of us.

I had been thinking about the possibility of developing a Taylor series expansion for X*X. another method is to do a Newton-Raphson iteration on a computer. I’m doubting if there is a closed form solution to this problem. I have been thinking about this problem for a long time and I was hoping that somebody would post a video about it, so THANK YOU for your video. Much appreciated

С таким же успехом можно сразу заметить, что 2

How about using log base 10 instead of log base e (because log base 10 of 10 = 1) ?

Very very nice. مرسی. I have been looking for a solution to this problem since last week and I posted it on my channel. Thank you for solving it

Where are the other solutions? The lambert W function has multiple branches.

Chat GPT came up with an approximate solution to this equation. Good enough for me.

Well, darnit, I can find the ln button on my calculator but I can't find that Lambert W button. How do I determine the value of W(ln 10)?

@pianoplayer123able

14 күн бұрын

There is a webiste called wolfram alpha. There you type it in.

did you every hear about tetration?

You'll have to explain the w() function, that looks as dodgy and true as limited developments... Whatever the Epsilon.

i ❤ Mathematics

I used to solve this type of equations graphically: ln x=ln 10/x...

This is fine, insofar as it goes. But there was no explanation of how to go from the Lambert W function to the approximate numeric value of X. This problem could more easily, and more understandably, be solved by using inspection and then simple iterative techniques to refine the approximation.

@DanDart

Ай бұрын

That would certainly give an approach. Not even big youtubers approximate manually, just use wolfram alpha, saying the function isn't elementary.

I can try out an initial value and then use the Newton-Raphson method xˣ = 10 Test x = 2 → 2² = 4 x = 3 → 3³ = 27 2 x = 2.5 → (2.5)^(2,5) ≈ 9.88 x = 2.6 → (2.6)^(2.6) ≈ 11.99 2.5 Newton-Rapshon (for f(x)=0) xₙ₋₁ = xₙ - (f(xₙ)/f '(xₙ)) xₙ₋₁ = approach xₙ = initial value f(xₙ) = function evaluated at initial value f '(xₙ) = derivative of the function evaluated at the initial value x ≈ xₙ = 2.5 f(x)=0 → xˣ -10 = 0 f '(x) = xˣ.(1+ln(x)) xₙ₋₁ = 2.5 - (((2.5)^(2.5)-10)/(((2.5)^(2.5))(1+ln(2.5)))) xₙ₋₁ ≅ 2.5062249694 x = xₙ₋₁ Solving in geogebra intersection between y=xˣ and y=10 I get the following value of x x ≈ 2.561841446 Comparing with the Newton-Raphson iterative numerical method, the approximation is acceptable. and if we want greater precision we can use the value of xₙ₋₁ as the initial value xₙ and repeat the iterative process of the method

@jellymath

Ай бұрын

oh no don't hurt me

OK But I think now the problem has multiplied, that is, I think the problem was not solved, but it was presented in a different language, but much more complicated than before. When it was solved, we would give the right side to the calculator and it would give us an approximate number. The problem is the same, only instead of writing x twice, we have written it only once, with the difference that even the answer can no longer be guessed subjectively and approximately, because we are talking about the inverse, exponential, Lambert and natural logarithm functions, while It used to be just an x to the power of x. I was hasty in the previous comment and now I understand that you were just being deceptive and I hope it was not intentional

solution as a special function? great... this way anything can be solved by simply adding yet another special function

@jellymath

Ай бұрын

you know, I guess the point of the Lambert W function is that it's easy to evaluate¿. In the end of the day, with this function you at least have a direct algorithm to finding the x out, how would you do it otherwise? But I agree, solutions with the W function aren't as satisfying

@silverhammer7779

Ай бұрын

Agreed. The presenter needed to show how he got from the W function to the approximate value of X. Here, he seems to be saying, "take it on faith that X is approximately 2.506" without any further explanation.

But what is x, if x^^x=10? And how do you calculate that?

@jellymath

Ай бұрын

oh my god

@MrMatthewliver

7 күн бұрын

Numerically, by substitution, with Excel.

@hansshattemvan8232

3 күн бұрын

@@MrMatthewliver Yes but how? What does that look like? And do you know approximately what the value of X is in this equation?

Je ne connais pas cette fonction w

Like most math videos this is way too long and wordy. People who are interested in this question aren't that slow.

@jellymath

Ай бұрын

I couldn't agree more. Crazy how he talks so much yet somehow says so little and almost nothing is happening

@davidwright8432

22 күн бұрын

Speak for yourself. Different folks, different styles of solving problems.

2.5061845 is the correct answer

@geeache1891

22 күн бұрын

2,5061841456