A Very Nice Math Olympiad Problem | Solve for n | Algebra

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24 окт 2024

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@cdiesch7000 4 дня назад

Very nice solution. But n^(1/n)=256^(1/256) has a second solution around 1.022. And the initial equation has even a third solution around -0.979 which got lost when puting the exponent of 1/n on both sides by which the negative was excluded. I have unfotunately no clue on how to calculate precisely these two other solutions.

@richardl6751 4 дня назад

I wrote a short program to calculate a slightly more precise value and came up with 1.022393 and -0.979017. Hope that helps.

@payoo_2674 3 дня назад

Use the Lambert W function W(■*e^■) = ■ n^32 = 2^n ln(n^32) = ln(2^n) 32*ln|n| = x*ln(n) ===> two cases 1st case: n > 0 32*ln(n) = n*ln(2) ln(n)*n^(-1) = ln(2)/32 ln(n)*(e^ln(n))^(-1) = ln(2)/32 ln(n)*e^(-ln(n)) = ln(2)/32 -ln(n)*e^(-ln(n)) = -ln(2)/32 W(-ln(n)*e^(-ln(n))) = W(-ln(2)/32) -ln(n) = W(-ln(2)/32) ln(n) = -W(-ln(2)/32) n = e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions n₁ = e^(-W₀(-ln(2)/32)) = 1.0223929402057803206527516798494005683768365119132864517728278977... in WolframAlpha: e^(-productlog(0,-ln(2)/32)) n₂ = e^(-W₋₁(-ln(2)/32)) = 256 # in WolframAlpha: e^(-productlog(-1,-ln(2)/32)) 2nd case: n < 0 32*ln(-n) = n*ln(2) ln(-n)*n^(-1) = ln(2)/32 -ln(-n)*n^(-1) = -ln(2)/32 ln(-n)*(-n)^(-1) = -ln(2)/32 ln(-n)*(e^ln(-n))^(-1) = -ln(2)/32 ln(-n)*e^(-ln(-n)) = -ln(2)/32 -ln(-n)*e^(-ln(-n)) = ln(2)/32 W(-ln(-n)*e^(-ln(-n))) = W(ln(2)/32) -ln(-n) = W(ln(2)/32) ln(-n) = -W(ln(2)/32) -n = e^(-W(ln(2)/32)) n = -e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution n₃ = -e^(-W₀(ln(2)/32)) = -0.979016934957784612322582550011650068748090048886011676265377083... in WolframAlpha: -e^(-productlog(0,ln(2)/32)) # e^(-W(-ln(2)/32)) = e^(-W(-8*ln(2)/(8*32))) = e^(-W(-ln(2^8)/256)) = e^(-W(-ln(256)*256^(-1))) = e^(-W(-ln(256)*(e^ln(256))^(-1))) = = e^(-W(-ln(256)*e^(-ln(256)))) = e^(-(-ln(256))) = e^ln(256) = 256