# Comparing powers with different bases without logarithms

I want to compare :

$17^{31}$ and $31^{17}$ , this is a solution

but I want another one and without using logarithms, only using the fact that

$17=16+1=(2^4)+1$ and $31=(2^5)-1$

how could it be done ?

• Do you care for another way which will not use neither logarithms nor the fact that you impose ? – Claude Leibovici May 11 '14 at 5:34
• please man go ahead – user2161721 May 11 '14 at 12:33

\begin{align}17 &= 2^4 +1\\ &> 2^4 \end{align} Hence \begin{align}17^{31} & > (2^4)^{31}\\ & = 2^{124} \end{align}
Meanwhile, \begin{align}31 &= 2^5 -1\\ &< 2^5 \end{align} and so, \begin{align}31^{17} &< (2^5)^{17}\\ &= 2^{85} \end{align}
In sum, $17^{31} > 2^{124} > 2^{85} > 31^{17}$.