It looks like you need to do a numerical solution for the Lambert W function evaluated for a real constant, so you might just as well do a numerical solution for the original equation. As an alternative solution, you should show how to differentiate the equation and use Newton's method to get a numerical solution.
take sqrt of both sides. after some exponent magic, you get sqrt(x)^sqrt(x) = sqrt(3). solve for a numerical solution for some value c where c = sqrt(x) and c^c = sqrt(3). then x = c^2
i think a new formula that can be develop from the lambert w function would be , If X.LogX = B , then X = e^W(B) , Solution for natural log variables equation
@@simonlevett4776 That should be obvious. For example, you have an equation, then you say you're going to divide by two and then write the equation again and write 2 in the demoninator, then you cancel the 2's and write the simplified eq. All you need to do is say "divide both sides by 2" - and write the result. The mathematics you are showing is far beyond the need to show every step. Saying a*b = b*a also does not need to be said or written.
It looks like you need to do a numerical solution for the Lambert W function evaluated for a real constant, so you might just as well do a numerical solution for the original equation.
As an alternative solution, you should show how to differentiate the equation and use Newton's method to get a numerical solution.
Tricky to see the right path. But practice makes the master. Thanks and greetings from Germany 😊
take sqrt of both sides. after some exponent magic, you get sqrt(x)^sqrt(x) = sqrt(3). solve for a numerical solution for some value c where c = sqrt(x) and c^c = sqrt(3). then x = c^2
It’s in my head.
i think a new formula that can be develop from the lambert w function would be , If X.LogX = B , then X = e^W(B) , Solution for natural log variables equation
Need solution without W function. This is will be genius.
It's probably impossible and may be provably impossible.
I havent seen the video yet, but i solved it by just thinking square root of x equals y, and then it gets pretty easy
Wychodząc od e^(sqrt(x)*lnx)=e^ln3 dostajemy rozwiązanie x=4[W(sqrt(3)/2)]^2 i jest prostsze!
I think you skipped a step. Around 6:50, how do you get from ln y = W(ln3/2) to just y = e^W(ln3/2) ?
By exponentiating both sides. If ln(a)=b then a = e^b
*a true genius
(x ➖ 3x+3).
I just looked at it and thought, the answer is a little more than 2.
I would like to see a solution with applied mathematics
x^(√x)=3
[(√x)²]^(√x)=3
Let n=√x
Domain: x>0
(n²)ⁿ=3
(nⁿ)²=3
|nⁿ|=√3
Reject negative value
nⁿ=√3
nⁿ=3^½
ln(nⁿ)=ln(3^½)
n[ln(n)]=½[ln(3)]
[ln(n)][e^(ln(n))]=½[ln(3)]
ln(n)=W(½[ln(3)])
n=e^W(½[ln(3)])
√x=e^W(½[ln(3)])
x=e^2W(½[ln(3)]) ❤
Too many intermediate steps.
How should it be done then ?
@@simonlevett4776he didn’t need to rewrite ln y * e ln y as e ln y * ln y for starters. That’s ridiculous…
@@simonlevett4776or repeating the simplest of power rules twice off to the side…
The important thing to know is the strategy. This gets lost by over-explaining every simple step.
@@simonlevett4776 That should be obvious. For example, you have an equation, then you say you're going to divide by two and then write the equation again and write 2 in the demoninator, then you cancel the 2's and write the simplified eq. All you need to do is say "divide both sides by 2" - and write the result. The mathematics you are showing is far beyond the need to show every step. Saying a*b = b*a also does not need to be said or written.
3 = x^√x
= (e^ln(x))^√x
= e^(ln(x) √x)
√x = x^(1/2) = e^(ln x)^(1/2) = e^(ln(x)/2)
3 = e^(ln(x) e^(ln(x)/2))
ln(3) = ln(x) e^(ln(x)/2)
ln(3)/2 = ln(x)/2 e^(ln(x)/2)
Recall definition of Lambert W:
u = W(u e^u)
ln(3)/2 = ln(x)/2 e^(ln(x)/2)
W(ln(3)/2) = W(ln(x)/2 e^(ln(x)/2))
= ln(x)/2
ln(x) = 2W(ln(3)/2)
x = e^(2W(ln(3)/2))
= 2.124799383022509033508588247984681694634624196480...
2.124799383022509033508588247984681694634624196480^√2.124799383022509033508588247984681694634624196480
= 2.999999999999999999999999999999999999999999999997...
Hiii
Sqrt(x)*ln(x)=ln(3), 1/2*ln(x) + ln(ln(x))=ln(ln(3)), with a=ln(x): 1/2*a+ln(a) =ln(ln(3)), a*exp(a/2)=ln(3), w(a/2*exp(a/2))=w(ln(3)/2), a/2=w(ln(3)/2), a=2*w(ln(3)/2), x=exp(2w(ln(3)/2)), right?
how to (do something) is not a question
This is not higher mathematics in any way. The maximum is the 8th grade of secondary school.
Why do you speak so raapidly????
let u=Vx , (u^2)^u=3 , 2u*lnu=ln3 , W(lnu*e^lnu)=W(ln3/2) , lnu=W(ln3/2) , u=e^(W(ln3/2)) , u=e^(0.376838694) ,
u=~ 1.45767 , recall u=Vx , Vx=~ 1.45767 , x=~ 2.1248 , test , 2.1248^1.45767=3 , same , OK ,