How I Solved An Equation with Floor Value | #numbertheory

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x^⌊x⌋ = 37
#floorequations #numbertheory #numbertheoryproblems
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[karl]

Before watching the video:
Since 3^3 < 37 and 4^4 > 37, x is strictly between 3 and 4, so floor(x) = 3, so x^floor(x) = x^3, so x = cuberoot(37)
..which is essentially roger7341's method, just a bit more verbal in description 😉

[elmer]

By inspection of expected base to power magnitude I set floor(x) = 3 and x = e^(ln(37)/3) = 3.33222...

[rob]

It's about 3.33222185164595
= 37^(1/3)