Math question for a "true" geniuses

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  • Опубліковано 29 жов 2024

КОМЕНТАРІ • 38

  • @dpasek1
    @dpasek1 2 місяці тому +7

    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.

  • @ganymed1236
    @ganymed1236 2 місяці тому +4

    Tricky to see the right path. But practice makes the master. Thanks and greetings from Germany 😊

  • @jwil4286
    @jwil4286 2 місяці тому +1

    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

  • @RyanLewis-Johnson-wq6xs
    @RyanLewis-Johnson-wq6xs 2 місяці тому +1

    It’s in my head.

  • @childrenofkoris
    @childrenofkoris 2 місяці тому +1

    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

  • @MadMax-um1fb
    @MadMax-um1fb 2 місяці тому +2

    Need solution without W function. This is will be genius.

    • @13thengineering33
      @13thengineering33 Місяць тому

      It's probably impossible and may be provably impossible.

  • @Netfriendship
    @Netfriendship 2 місяці тому +2

    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

  • @Zbigniew-b3u
    @Zbigniew-b3u 2 місяці тому

    Wychodząc od e^(sqrt(x)*lnx)=e^ln3 dostajemy rozwiązanie x=4[W(sqrt(3)/2)]^2 i jest prostsze!

  • @chrisjust7445
    @chrisjust7445 Місяць тому

    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) ?

    • @azmath2059
      @azmath2059 2 дні тому

      By exponentiating both sides. If ln(a)=b then a = e^b

  • @LucasF.-sw7ep
    @LucasF.-sw7ep 2 місяці тому

    *a true genius

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 2 місяці тому

    (x ➖ 3x+3).

  • @jegog.
    @jegog. Місяць тому

    I just looked at it and thought, the answer is a little more than 2.

  • @mia_foni
    @mia_foni 2 місяці тому

    I would like to see a solution with applied mathematics

  • @ChavoMysterio
    @ChavoMysterio 2 місяці тому +2

    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)]) ❤

  • @NoferTrunions
    @NoferTrunions 2 місяці тому +11

    Too many intermediate steps.

    • @simonlevett4776
      @simonlevett4776 2 місяці тому

      How should it be done then ?

    • @ChrisSpille
      @ChrisSpille 2 місяці тому +3

      @@simonlevett4776he didn’t need to rewrite ln y * e ln y as e ln y * ln y for starters. That’s ridiculous…

    • @ChrisSpille
      @ChrisSpille 2 місяці тому +2

      @@simonlevett4776or repeating the simplest of power rules twice off to the side…

    • @musicsubicandcebu1774
      @musicsubicandcebu1774 2 місяці тому +5

      The important thing to know is the strategy. This gets lost by over-explaining every simple step.

    • @NoferTrunions
      @NoferTrunions 2 місяці тому +1

      @@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.

  • @edwardhuff4727
    @edwardhuff4727 2 місяці тому +1

    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...

  • @Iitian2025-t7g
    @Iitian2025-t7g 2 місяці тому +1

    Hiii

  • @jenskluge7188
    @jenskluge7188 2 місяці тому

    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?

  • @dbtx
    @dbtx 2 місяці тому

    how to (do something) is not a question

  • @olegdmitriev6854
    @olegdmitriev6854 2 місяці тому +1

    This is not higher mathematics in any way. The maximum is the 8th grade of secondary school.

  • @georgelaing2578
    @georgelaing2578 2 місяці тому

    Why do you speak so raapidly????

  • @prollysine
    @prollysine 2 місяці тому

    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 ,