exponential

$(2+1)^{98}>2^{98}+98*2^{97}*1=2^{97}(2+98)=2^{97}*100$

$2^{97}*100>2^{97}*50>2^{97}*2^{5}=2^{102}$

$3^{98}>2^{102}>2^{100}$

Therefore, $3^{98}>2^{100}$

2^100 or 3^98 ?

2^100/100 or 3^98/100 ?

2^1 or 3^0.98 ?

0.98 is approximately equal to 1

2^1 or 3^1 ?

2^1 < 3^1

3^98 > 2^100