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 ?
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$\begingroup$ Do you care for another way which will not use neither logarithms nor the fact that you impose ? $\endgroup$ – Claude Leibovici May 11 '14 at 5:34
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$\begingroup$ please man go ahead $\endgroup$ – user2161721 May 11 '14 at 12:33
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$$\begin{align}17 &= 2^4 +1\\ &> 2^4 \end{align}$$ 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, $17^{31} > 2^{124} > 2^{85} > 31^{17}$.
