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what is the largest number here? how to find these with out inspection but a proper mathematical rule?

$$\Large 2^{3^4},2^{4^3},3^{2^4},3^{4^2},4^{2^3},4^{3^2}$$

Thank you

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    $\begingroup$ Should the answer be written on papyrus or parchment? $\endgroup$ – user64494 Sep 14 '13 at 17:13
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As $\displaystyle a^{m^n}=a^{(m^n)}$ and $\displaystyle (a^m)^n=a^{m\cdot n}$

$2^{3^4}=2^{81}$

$2^{4^3}=2^{64}$

$3^{2^4}=3^{16}$

$3^{4^2}=3^{16}$

$4^{2^3}=(2^2)^{(2^3)}=(2^2)^8=2^{16}$

$4^{3^2}=(2^2)^{(3^2)}=(2^2)^9=2^{18}$

As we are interested only in finding the largest, we can safely consider $2^{81}$ and $3^{16}$ (why?)

$2^{81}=2\cdot(2^5)^{16}>(32)^{16}>3^{16}$

Also, $2^4>3, (2^4)^{16}>3^{16}\implies 2^{64}>3^{16}$

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    $\begingroup$ (+1) Alternatively $2^{81} > 4^{40} > 3^{16}$. $\endgroup$ – 6005 Sep 14 '13 at 17:06

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