Which of the following powers is bigger: $2^{41}$ or $3^{24}$?

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

$$ 2^{41} = 2,199,023,255,552 > 282,429,536,481 = 3^{24} $$

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.

Answer 5

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 $$ 2^{41} > 2^{40} = (2^{10})^4 > (10^3)^4 = 10^{12} > (3^2)^{12} = 3^{24}. $$

Answer 6

A definitive answer using log functions:

log(3) / log(2) = 1.58496

3²⁴ = (2^1.58496)²⁴ = 2^{1.58496 * 24} = 2^38.039

Therefore: 2⁴¹ > 3²⁴