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

Question

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?

Answer 1

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 $3^6=729>512=2^9$.

Answer 2

$6^{16}=2^{16} \cdot 3^{16}$ 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 $k=4$.