Talk:Equality of Integers to the Power of Each Other

Possible solution
$ n^m = m^n \implies m \log_n n = n \log_n m \implies m = n \log_n m $

Assume $ m > n $.

Since $ m, n \in \N $, $ \log_n m \in \N $ if $ m = n^k $, where $ k \in \N $ and $ k > 1 $ by assumption, but otherwise a free parameter.

Hence, $ n^k = k n $.

For $ n \ne 0 $ the solution reads


 * $ \displaystyle n = k^\frac{ 1 }{ k - 1 } $

Define $ t = k - 1 $.

$ \lim_{ k \to 1 } n = e $

$ \lim_{ k \to \infty } n = 1 $

To check for intermediate maximum consider the first derivative:


 * $ \displaystyle \frac{ \mathrm d n }{ \mathrm d k } = \frac{ k^{ \frac{ 1 }{ k -1} } }{ k - 1 } \left ( { \frac{ 1 }{ k } - \frac{ \ln k}{ k - 1 } } \right ) $

Our desired solution constraints the prefactor to be positive.

The term in brackets vanishes only for $ k = 1$, hence for $ k > 1 $ there is no extremum, $ n \left ( { k } \right ) $ is monotonically decreasing, and $ 1 < n < e $.

The only natural solution is $ n = 2 $

The only $ k $ that satisfies this is $ k = 2 $.

Therefore, $ n = 2 $, $ m = 2^2 = 4 $. Now assume $ m < n $ and repeat the calculation. --Julius (talk) 09:03, 18 May 2017 (EDT)