An Exponent That n-tuples | Problem 474

[@scottleung9587]

Cool!

[@krabbediem]

ni?! Be very careful, the knights may be lurking

[@aplusbi]

😄

[@Don-Ensley]

problem i ᶻ = n i Assume n is integer. Take the natural logarithm of each side. Bring down exponents. z ln i = ln n + ln i Subtract ln i from each side. z ln i - ln i = ln n Factor. z ln i - ln i = ln n ( z - 1 ) ln i = ln n Substitute for ln i. ln i = i π ( 1/2 + 2 K ), K an integer. ( z - 1 ) i π ( 1/2 + 2 K ) = ln n Multiply by -i. ( z - 1 ) π ( 1/2 + 2 K ) = - i ln n Divide by π ( 1/2 + 2 K ). ( z - 1 ) = - i ln n / [ π ( 1/2 + 2 K ) ] z = 1 - 2 i ln n / [ π ( 4 K + 1 ) ] answer z ∈ { 1 - 2 i ln n / [ π ( 4 K + 1 ) ], ( n, K ∈ ℤ ) }

[@davidseed2939]

Can we not go directly from I^z=ni The very end Zlni= ln(n) +lni z= ln(n)/ln(i) +1 z= 1+ ln(n)/( ia π )) z=1 -i ln(n)/(a.π) where a= 2k+1/2

[@mcwulf25]

At the end, you used k on both sides. Then cancelled. This is wrong. Mostly. You should have not 1 but (1 + 4k)/(1 + 4m)

[@aplusbi]

Does that work?

[@mcwulf25]

@aplusbi Should do. We can't simply assume k=m.