An algebra problem by Ayush G Rai

Since they are all positive, taking $100^\text{th}$ root preserves their ordering.
This means that we only need to compute $4^3,5^2,3^4,2^5$ . They are $64,25,81,32$ respectively.
The largest is $81$ , corresponding to the choice $3^{400}$ .

$4^{300}$ $= (4^3)^{100}$

$3^{400}$ $= (3^4)^{100}$

$2^{500}$ $= (2^5)^{100}$

$5^{200}$ $= (5^2)^{100}$

We know $5^2 < 2^5 < 4^3 < 3^4$

$\large →5^{200} <2^{500}< 4^{300} <3^{400}$

Transform all of tha numbers to powers of 100, then you will see 3^4 is bigger

nice solutions!!!