• Solve x for xx*x^x = 2
  • Note that the Lambert W function W(x) is the inverse of f(x) = xex
  • zkfcfbzr@lemmy.worldEnglish
    4·
    2 months ago
    solution

    x^(x*x^x) = 2

    (x^x)^(x^x) = 2

    k = x^x

    k^k = 2

    k*ln(k) = ln(2) → Log of both sides

    ln(k) * e^ln(k) = ln(2) → k = e^ln(k)

    f(ln(k)) = ln(2)

    ln(k) = W(ln(2))

    ln(x^x) = W(ln(2))

    ln(x)*e^ln(x) = W(ln(2)) → Same step as noted earlier

    f(ln(x)) = W(ln(2))

    ln(x) = W(W(ln(2))

    x = e^W(W(ln(2)))

    x ≈ 1.3799703966 (via Wolfram|Alpha, seems to be the correct value)

  • siriusmart@lemmy.worldOPM
    3·
    2 months ago

    This is a genuinely fun one

    hint 1 (don’t peek unless you really need it)

    spoiler

    try solving a simpler version of the problem

    hint 2 (supplementary to hint 1)

    spoiler

    the simpler version is xx=2

    hint 3 (unrelated to 1 and 2)

    spoiler

    xx*x^x=(xx)^(xx)

    hint 4 (its basically the answer)

    spoiler

    xx=e^(ln x)(e^(ln x))

    Solution: https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd solutions/2024-05-05_fake-power-tower.html

    spoiler

  • Limonene@lemmy.world
    21·
    2 months ago

    The text of this post appears wrong on old.lemmy.world. It says “Solve x for x^x*x^x^ = 2” with no superscripts. It appears correctly on lemmy.world.

    I assume we’re meant to find an expression of W() and square roots and stuff, which expresses an exact answer. Since finding a decimal approximation somewhere between 1 and 2 using a binary search would be too easy.

    • Limonene@lemmy.world
      31·
      2 months ago

      I believe it is:

      spoiler

      e^W(W(ln(2))

      spoiler 
      x=W(x)*e^(W(x))

x^(x*x^x)=2
x*x^x*ln(x)=ln(2)
x*e^(ln(x)*x)*ln(x)=ln(2)
u=x*ln(x)
u*e^u=ln(2)
u=W(ln(2))
x*ln(x)=W(ln(2))
e^(ln(x)*x)=e^W(ln(2))
x^x=e^W(ln(2))
x = square-super-root(e^W(ln(2)))
wikipedia says this is equivalent to:
x=e^W(ln(e^W(ln(2))))
but I don't know how they arrive at that.
x=e^W(W(ln(2))

working backwards to verify:
x=e^W(W(ln(2))
ln(x)=W(W(ln(2))
ln(x)*x=W(ln(2))
ln(x)*x*e^(ln(x)*x)=ln(2)
ln(x)*x*x^x=ln(2)
e^(ln(x)*x*x^x)=2
x^(x*x^x)=2
      • siriusmart@lemmy.worldOPM
        2·
        2 months ago

        x^x=e^W(ln(2)) isn’t wrong, but it’s in a form that’s inconvenient to say the least.

        Picking up from x*ln(x)=W(ln(2))

        spoiler

        spoiler

        x^x is a far superior substitution, but it takes a bit to notice it