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An Exponent That Triples | Problem 380 

aplusbi
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6 окт 2024

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Комментарии : 12   
@mcwulf25
@mcwulf25 6 дней назад
If n=1 you have 5pi at the bottom.
@trojanleo123
@trojanleo123 5 дней назад
z = (4m+1)/(4k+1) - i[2.ln3/π(4k+1)] For principle values m=k=0 z = 1 - i*2ln3/π
@scottleung9587
@scottleung9587 6 дней назад
Cool!
@lauciansylvaranth2285
@lauciansylvaranth2285 6 дней назад
Maybe do an exponent that multiplies? Like, can you do: W^Z =nW? or at least i^Z = ni? I am a fan of general cases, you see?
@aplusbi
@aplusbi 6 дней назад
Great idea! 😍
@dorkmania
@dorkmania 6 дней назад
Dividing i^(4m + z) = 3•i^(4n + 1) by i^(4n + 1) i^(4m + z - 4n - 1) = 3 => i^(4(m - n) + z - 1) = 3 Or i^(4k+ z - 1) = 3, where k = m - n
@viniaz2997
@viniaz2997 5 дней назад
Did you do i to the power of z equals negative i? e to the power of z equals negative e?
@aplusbi
@aplusbi 5 дней назад
sounds like great ideas!
@trojanleo123
@trojanleo123 5 дней назад
Ooooh. That should be fun!!!
@mcwulf25
@mcwulf25 6 дней назад
Why wouldn't you need n? Isn’t it integral to the substitution for i?
@aplusbi
@aplusbi 6 дней назад
I think because z is a variable and having multiple values for 3i on the right hand side provides all the possible solutions. I have a feeling only certain n values will work. There is always an exponent z that makes this equation possible without considering the multiple representations of i at the base. Using n breaks that association. This is my guess btw
@Don-Ensley
@Don-Ensley 5 дней назад
z = 1 - i (2 ln 3)/(4K +1) , K∈ℤ
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