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Calculate the maximum between the given numbers $$ \max(2^{41},3^{24})\text{.}$$ I got stuck when I tried to decompose the exponent as $41$ is a prime number.

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$$2^{41}\gt 2^{40}=(2^5)^8\gt (3^3)^8=3^{24}$$

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  • $\begingroup$ Should I know 2^5 is bigger than 3^3 without calculating that 2^5=32 and 3^3=27? $\endgroup$ – Frank Shmrank Nov 2 '15 at 19:29
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    $\begingroup$ @Dean MacGregor, no, except that many people know those small powers by rote (e.g., binary sequence in computer science and perfect cubes to reduce cube roots). $\endgroup$ – Daniel R. Collins Nov 2 '15 at 20:51
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Recall that $2^{10} = 1024 > 1000 = 10^3$ (well-known from computer science as the size of a kilobyte), and $3^2 = 9 < 10$, so that $$ 2^{41} > 2^{40} = (2^{10})^4 > (10^3)^4 = 10^{12} > (3^2)^{12} = 3^{24}. $$

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It suffices to see $2^{20}>3^{12}\iff 2^{10}>3^6\iff 2^{5}>3^3$

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Alternatively $$2^{41} > 2^{40} = (2^8)^5 > (3^5)^5 = 3^{25}$$ shows an even stronger statement.

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A definitive answer using log functions:

log(3) / log(2) = 1.58496

324 = (21.58496)24 = 21.58496 * 24 = 238.039

Therefore: 241 > 324

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  • $\begingroup$ instead of calculating $log 3 / log 2$, you can simply calc $log_2 3$ ... binary log is quite common IMO, no need to go for natural one here. $\endgroup$ – vaxquis Nov 3 '15 at 13:13
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$$ 2^{41} = 2,199,023,255,552 > 282,429,536,481 = 3^{24} $$

Edit

Apparently this answer isn't 'mathematical' enough for some commenters as it is direct, simple, and verifiable, but doesn't use log functions and all that other Algebra II knowledge we are so proud of. Anyone in the world should be able to make the above calculation to more than the required degree of accuracy, as follows:

$$ 2^{41} = 2 * (2^{10})^4 = 2 * (1024)^4 \underset{\approx}{>} 2 * (1000)^4 = 2,000,000,000,000 $$ and $$ 3^{24} = (3^2)^{12} = 9^{12} \underset{\approx}{<} 10^{12} = 1,000,000,000,000 $$

Edit 2

@AriBrodsky uses exactly the above calculation in an answer that was posted after mine but before my edit. I didn't see it until now. It is clearly the best answer, but I leave my edit for clarity.

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  • $\begingroup$ you made a mistake when calculating $2^{41}$ $\endgroup$ – Jorge Fernández Hidalgo Nov 2 '15 at 17:37
  • $\begingroup$ @dREaM 2199023255552? $\endgroup$ – Tim Nov 2 '15 at 18:24
  • $\begingroup$ yup${}{}{}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Nov 2 '15 at 18:26
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    $\begingroup$ @jwg IMO the obvious implication of the "Calculate (...)" question was "find a way to get the result of $max$ without doing the actual numerical evaluation" $\endgroup$ – vaxquis Nov 2 '15 at 23:38
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    $\begingroup$ @vaxquis We don't agree about the 'obvious' meaning of the word 'Calculate' then. $\endgroup$ – jwg Nov 3 '15 at 6:48

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