Do You Like Powers?

$2^{5555} = 32^{1111}$
$3^{3333} = 27^{1111}$
$6^{2222} = 36^{1111}$

Since $27 < 32 < 36$ , the required order of the three numbers is

${ 3 }^{ 3333 }$ , ${ 2 }^{ 5555 }$ , ${ 6 }^{ 2222 }$

$\Large A=(2^5)^{1111}, ~~B=(3^3)^{1111},~~C=(6^2)^{1111}\\ 2^5=32,~~3^3=27, 6^2=36.\\ \therefore~~B<A<C.$

In this entire solution, all the $\log$ functions are in base-10.

We know that $\log(2) = 0.301 , \log(3) = 0.477$ .

$\log(A) = 5555 \log(2) = 5555 \times 0.301 \approx 1672.2$

$\log(B) = 3333 \log(3) = 3333 \times 0.477 \approx 1590.2$

$\log(C) = 2222 \log(6) = 2222( \log(2) + \log(3) ) = 2222(0.301+0.477) \approx 1729.1$

Since we can see that $\log(C) > \log(A) > \log(B)$ and since $A,B,C > 0$ , we conclude that $C>A>B$ .

A has 1673 digits.

B has 1591 digits.

C has 1730 digits.

Clearly C>A>B

2^5 = 32; 3^3= 27; 6^2 = 36; 27 < 32< 36 So, B<A<C.

The screen capture is only 6^2

First we can consider only one digit of power. Then calculate, 2^5=32; 3^3=27; 6^2=36.So B<A<C.

it's some what logical question: 2^5= 32 ; 3^3= 27 ; 6^2=36

We can simply look for assessment of A=3^2 ,B=3^3. c=6^2 that give us clue that C is lowest,A is more tha C and B is biggest ,therefore B <A<C

2^(5)1111 3^(3)1111 & 6^(2)*1111 Therefore, 32^1111 27^1111 36^1111 ,i jus focused on thr values of base,i don't mind the exponents coz they are just similar :D

Bring all the number to the power of 1111 and then compare .