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

[@cdiesch7000]

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]

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]

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

[@RamendraPaswan-f2k]

n^1/n=2^1/32
n^1/n=2^1.8/32.8
n^1/n=(2^8)^1/256
n^1/n=(256)^1/256
n=256

[@SpencersAcademy]

Nice one! 👏