Determining the relative size of $a^n$ and $b^m$ without using logarithms Example, which is larger $ 17^{105} $ vs $ 31^{84} $? Make the deternimination without resorting to logs, or Excel either.
 A: HINT:
$$(84,105)=21$$
So, it is sufficient to check $17^5<=>31^4$ 
$31^2=961,17^2=289$  
Now $\dfrac{961}{289}<4\implies31^4<17^4\cdot4^2<17^4\cdot17=17^5$
A: $$ (16 + 1)^{105} \ = \  (2^4 + 1)^{105} \ = \ 2^{420} \ + \ 105 \cdot \ 2^{416} \ + \ \ldots  $$
versus
$$ (32 - 1)^{84} \ = \ (2^5 - 1)^{84} \ = \ 2^{420} \ - \ 84 \cdot \ 2^{415} \ + \ \ldots $$
Note that all of the "binomial terms" for $ \ 17^{105} \ $ are positive, while those for $ \ 31^{84} \ $ alternate signs. (EDIT: Also, since the exponent for $ \ 32^{84} \ $ is lower, the binomial coefficients in its expansion are smaller, so there is no possibility that the positive terms will somehow "catch up" with, much less surpass, those for $ \ 17^{105} \ $ .)
A: We know that $$ 2\cdot17 > 31$$ 
$$\implies (2\cdot17)^{84} > 31^{84} $$
$$\implies 2^{84}\cdot17^{84} > 31^{84} $$
Since $$105-84=21$$
And $$2^4<17$$
$$\implies (2^4)^{21}<17^{21}$$
$$\implies 2^{84}<17^{21}$$
Clearly $$17^{21}\cdot17^{84} > 31^{84}$$
$$\implies 17^{105} > 31^{84}$$
