Which of the following powers is bigger: $2^{41}$ or $3^{24}$?
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.
Solution 1:
$$2^{41}\gt 2^{40}=(2^5)^8\gt (3^3)^8=3^{24}$$
Solution 2:
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 $$ 2^{41} > 2^{40} = (2^{10})^4 > (10^3)^4 = 10^{12} > (3^2)^{12} = 3^{24}. $$