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Which is the "fastest" paper-pencil method to compare (find which one is greater) $\sqrt[17]{6}$ and $\sqrt[16]{4}$?

My analysis bought this whole thing down to comparing which is greater between $6^{16}$ and $2^{34}$.Then by using the technique mention here doesn't gives the precise answer as $\log_2 6$ lies between $2 \text { and }3$, how else I could do this?

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    This is another [related](http://math.stackexchange.com/questions/15928/how-to-compare-powers-without-calculating) discussion.2011-12-06

2 Answers 2

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You want to know whether $2^{1/17} 3^{1/17} \geq 2^{1/8}$. Raising both sides to appropriate powers and cancelling, this is equivalent to $3^8 \geq 2^9$. But $3^6=729>512=2^9$.

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    I don't know why can't I see that before... thanks anyways.2011-12-06
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$6^{16}=2^{16} \cdot 3^{16}$ and so you only need to compare $3^{16}$ and $2^{18}$. Now $3^{12}>2^{12}$ and so you need to compare $3^{4}=81$ and $2^{6}=64$. For this last step, the question is: when is $3^k > 2^{k+2}$? The smallest solution is $k=4$.

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    Note that you were rather lucky here, since sour simplification (using 3^{12} > 2^{12}) wouldn't always yield a result. Consider for example $3^{13}$ and $2^{14}$ : 3^1 < 2^2 , and yet 3^{13} > 2^{14}. Thus writing "_Now 3^{12}>2^{12} and so you need to compare (...)_" is not really rigorous (it implies it is necessary, while it is just sufficient); instead I think you should write something like "_Now 3^{12}>2^{12} and so proving 3^{4}=81 > 2^{6}=64 would be sufficient_".2012-08-25