finding the second smallest number

Question

Please tell me how to solve this kind of problem in a fast manner.

Which of the following is the second smallest number: $2^{120}$, $3^{80}$ and $10^{30}$?

Hope you could show me the best solution, thanks.

Answer 1

These three numbers have the same ordering as their tenth roots, which are $2^{12}$, $3^8$, and $10^3$; all of these are easily calculated by hand. The hardest is $3^8$, and it’s pretty easy:

$$3^8=9^4=81^2=6561\;,$$

all of which was done in my head.

Answer 2

Use the facts that $2^{10} = 1024 \geq 1000 = 10^3$

and $3^2 = 9 < 10^1$

and $2^3=8< 3^2$

so $2^{120}=\left(2^{10}\right)^{12} > \left(10^3\right)^{12}=10^{36} > 10^{30}$

and $2^{120}=\left(2^3\right)^{40} < \left(3^2\right)^{40} = 3^{80}$