# Which is the “fastest” paper-pencil method to compare $\sqrt[17]{6}$ and $\sqrt[16]{4}$?

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 discussion. –  Quixotic Dec 6 '11 at 10:37

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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What are the appropriate powers you raised and cancel? –  Quixotic Dec 6 '11 at 11:19
raise both sides to the power $8 \cdot 17$, then cancel $2^8$ from both sides –  mt_ Dec 6 '11 at 11:24
I don't know why can't I see that before... thanks anyways. –  Quixotic Dec 6 '11 at 11:27
$6^{16}=2^{16} \cdot 3^{16}$ 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 $k=4$.
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". –  Clément Aug 25 '12 at 9:37