[2024/05/05] Fake power tower

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)

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

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hint 2 (supplementary to hint 1)

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the simpler version is x^x=2

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hint 3 (unrelated to 1 and 2)

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x^{x*x^x}=(x^x)^(x^x)

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hint 4 (its basically the answer)

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

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Solution: https://gmtex.siri.sh/fs/1/School/Extra/Maths/Qotd solutions/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.