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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)
ggs
This is a genuinely fun one
hint 1 (don't peek unless you really need it)
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try solving a simpler version of the problem
hint 2 (supplementary to hint 1)
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the simpler version is x^x^=2
hint 3 (unrelated to 1 and 2)
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x^x*x^x^=(x^x^)^(x^x^)
hint 4 (its basically the answer)
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x^x^=e^(ln x)(e^(ln x))
Solution: https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd%20solutions/2024-05-05_fake-power-tower.html
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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.
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
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