• Solve x for xx*x^x = 2
  • Note that the Lambert W function W(x) is the inverse of f(x) = xex
  • @Limonene
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    26 months ago

    I believe it is:

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    e^W(W(ln(2))

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    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
    
    • @siriusmartOPM
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      26 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))

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